Answer:
a) The electron mobility for this metal is 0.00455 m²V/s
b) There are 4.12 * 10^28 free electrons per cubic meter
Explanation:
Step 1: Data given
electrical resistivity of 3.3 * 10^−8(Ω⋅m)
A current of 25 A is passed through a specimen of this metal 15 mm thick.
A magnetic field of 0.95 tesla.
A Hall voltage of −2.4×10−7V is measured
Step 2: Calculate the conductivity
σ= 1/ ρ
⇒ with ρ = electrical resistivity of 3.3*10^−8Ω*m
σ = 1/3.3*10^−8
σ = 30303030.3 = 3*10^7/ Ω*m
step 3: Calculate the electron mobility
µe = Vdσ / IxBz
⇒ with µe = the electron mobility in m²V/s
⇒ V = voltage of 2.4*10^−7V
⇒ d = 0.015 meter
⇒ σ = the conductivity = 3*10^7/ Ω*m
⇒ Ix = the current = 25 A
⇒ Bz = magnetic field = 0.95 tesla
µe = (2.4 * 10^-7 * 0.015 * 3*10^7) / (25*0.95)
µe = 0.00455 m²V¨/s
The electron mobility is 0.00455 m²V¨/s
Step 4: Calculate the number of free electrons per cubic meter
n = σ/e*µe
⇒ e= the charge of an electron = 1.6 * 10^-19 Coulombs
⇒ σ = The conductivity = 3*10^7/ Ω*m
⇒ µe = The electron mobility = 0.00455 m²V¨/s
n = 3*10^7/ Ω*m /(1.6 * 10^-19 * 0.00455)
n = 4.12 * 10^28 / m³
There are 4.12 * 10^28 free electrons per cubic meter
The electron mobility of the metal is 3.18 × 10^-3 m^2/Vs. The number of free electrons per cubic meter is 1.52 × 10^29 electrons/m^3.
Explanation:To calculate the electron mobility and the number of free electrons per cubic meter for the given metal, we can use the equations involving Hall effect and resistivity. The formula for electron mobility is given by μ = -VH / (B * I * t), where VH is the Hall voltage, B is the magnetic field, I is the current, and t is the thickness. Substituting the values, we get the electron mobility as 3.18 × 10^-3 m^2/Vs.
The formula for the number of free electrons per cubic meter is given by n = 1 / (e * R), where e is the charge of an electron (1.6 × 10^-19 C) and R is the resistivity. Substituting the values, we get the number of free electrons per cubic meter as 1.52 × 10^29 electrons/m^3.
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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.70kg of water at 30.5 C. During the reaction 97.1 kJ of heat flows out of the flask and into the bath.
Calculate the new temperature of the water bath. You can assume that the specific heat capacity of water under these conditions is 4.18J / g*K. Round to the 3 significant digit.
Answer:
The new temperature of the water bath is 36.8°C
Explanation:
A Calorimety problem to apply this formula:
Q = m . C . ΔT
Where Q is heat
m is mass
C is specific heat capacity
ΔT = T° final - T° initial
Let's make some conversion before
97.1 kJ .1000 = 97100 Joules
3.70 kg . 1000 = 3700 g
This because the specific heat capacity units
ΔT it's a difference and mathematically, in numbers, it is the same value in °C as in K
97100 J = 3700 g . 4.18J / g°C ( T°final - 30.5°C)
97100 J = 15466 J/°C (T°final - 30.5°C)
97100 J / 15466 °C/J = (T°final - 30.5°C)
6.28°C = T °final - 30.5°C
6.28° C + 30.5°C = T°final
36.8°C = T° final
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g*K.
Answer: 54.4 kJ/mol
Explanation:
First we have to calculate the moles of HCl and NaOH.
[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole[/tex]
The balanced chemical reaction will be,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH
Thus, the number of neutralized moles = 0.05 mole
Now we have to calculate the mass of water:
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = [tex]50ml+50ml=100ml[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 100 g
[tex]T_{final}[/tex] = final temperature of water = [tex]27.5^0C[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = [tex]21.0^0C[/tex]
Now put all the given values in the above formula, we get:
[tex]q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C[/tex]
[tex]q=2719.6J=2.72kJ[/tex]
Thus, the heat released during the neutralization = 2.72 KJ
Now we have to calculate the enthalpy of neutralization per mole of [tex]HCl[/tex]:
0.05 moles of [tex]HCl[/tex] releases heat = 2.72 KJ
1 mole of [tex]HCl[/tex] releases heat =[tex]\frac{2.72}{0.05}\times 1=54.4KJ[/tex]
Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ
Answer:
54.34 kJ/mol of HCl
Explanation:
The calorimeter is a device used to determine the heat that is lost or gained, by a reaction. When the temperature change without a phase change, the heat (Q) can be calculated by:
Q = m*c*ΔT
Where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature variation (final - initial). The mass of the solution is the density multiplied by the volume:
m = 1.0 g/mL * 100 mL = 100 g
The temperature variation in °C is equal to the temperature variation in K, thus:
Q = 100g * 4.18J/gK * (27.5 - 21.0)K
Q = 2717 J
Thus, the solution gained 2717 J of heat. The enthalpy is how much of this energy is inside the matter, thus, it is the heat divided by the number of moles of a substance.
The number of moles of HCl is the volume (50 mL = 0.05 L) multiplied by the concentration:
n = 0.05 L * 1.0 M = 0.05 mol
The enthalpy is the heat divided by the number of moles:
H = 2717/0.05
H = 54340 J/mol of HCl
H = 54.34 kJ/mol of HCl
Which of the slightly soluble salts below will be more soluble in acidic solution than in pure water?
a. Mg(OH)2
b. CaCO3
c. AgCl Mg(OH)2
d. only Mg(OH)2,
e. CaCO3, and
f. AgCl
g. CaCO3 only
h. both Mg(OH)2 and CaCO3
Answer:
h. both Mg(OH)₂ and CaCO₃
Explanation:
Let's consider the solution of Mg(OH)₂ according to the following equation:
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
In acidic solution, OH⁻ reacts with H⁺ to form H₂O.
OH⁻(aq) + H⁺(aq) ⇄ H₂O(l)
According to Le Chatelier's principle, since [OH⁻] decreases, the solution of Mg(OH)₂(s) shifts toward the right, increasing its solubility.
Let's consider the solution of CaCO₃ according to the following equation:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
In acidic solution, CO₃²⁻ reacts with H⁺ to form HCO₃⁻.
CO₃²⁻(aq) + H⁺(aq) ⇄ HCO₃⁻(aq)
According to Le Chatelier's principle, since [CO₃²⁻] decreases, the solution of CaCO₃(s) shifts toward the right, increasing its solubility.
Let's consider the solution of AgCl according to the following equation:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Cl⁻ does not react with H⁺ because it comes from a strong acid (HCl). Therefore, the solubility of AgCl(s) is not affected by the pH.
Mg(OH)₂ and CaCO₃ are more soluble in acidic solutions due to reactions with H+ ions that remove the OH- and CO₃²⁻ from the solution, driving the dissolution equilibrium forward.
Explanation:The question asks which of the slightly soluble salts listed will be more soluble in acidic solution than in pure water. Specifically, salts like Mg(OH)₂ (magnesium hydroxide) and CaCO₃ (calcium carbonate) will be more soluble in an acidic solution. This is because the acid in the solution will react with the anionic part of the salt, which in the case of Mg(OH)2 is OH- and for CaCO₃ is CO₃²⁻. For example, in an acidic solution, H+ ions will react with OH- to form water, which effectively removes OH- from the solution and drives the dissolution equilibrium forward, increasing the solubility of Mg(OH)₂. Similarly, H+ ions will react with CO₃²⁻ to form HCO₃- (bicarbonate) or even further to H₂CO₃ (carbonic acid), which are more soluble than the carbonate ion, hence increasing the solubility of CaCO₃.
As for AgCl (silver chloride), it will also be more soluble in acidic solution because the chloride ion is not basic, and it does not react with H+ to form a weaker acid. Therefore, the correct answers that are more soluble in acidic solution than in pure water are Mg(OH)₂ and CaCO₃.
Phosphorus is present in seawater to the extent of 0.07 ppm by mass. You may want to reference (Page) Section 18.3 while completing this problem. Part A Assuming that the phosphorus is present as dihydrogenphosphate, H2PO43−, calculate the corresponding molar concentration of phosphate in seawater. The density of seawater is 1.025 g/mL.
The molar concentration of phosphate (present as dihydrogen phosphate) in seawater can be calculated by converting the given mass concentration of phosphate (0.07 ppm) into moles using the molar mass of phosphate. After this conversion, the phosphate concentration is approximately 7.14 x 10^-4 mol/L.
Explanation:The question asks us to calculate the molar concentration of the dihydrogen phosphate ion in seawater. Remember that molar concentration is the amount of a solute (in this case phosphorus as dihydrogen phosphate) divided by the volume of the solution (seawater). We are given that the concentration of phosphorus is 0.07 ppm by mass, which, converted into grams per liter (g/L), yields 0.07 g/L. Additionally, the molar mass of H₂PO₄⁻ is approximately 97.994 g/mol.
Step-by-step calculation:
First, convert the given mass concentration of phosphorus into moles by using the molar mass of phosphorus: (0.07 g/L) / (97.994 g/mol) = 7.14 x 10⁻⁴ mol/L.Since the phosphorus was assumed to be in the form of dihydrogen phosphate, we don't need any further calculations as the molar concentration of dihydrogen phosphate is equal to that of phosphorus.Therefore, the molar concentration of phosphate in seawater is approximately 7.14 x 10⁻⁴ mol/L.
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5 Fe2+ + MnO4− + 8 H+ ⇄ 5 Fe3+ + Mn2+ + 4 H2O In a titration experiment based on the equation above, 25.0 milliliters of an acidified Fe2+ solution requires 14.0 milliliters of standard 0.050-molar MnO4− solution to reach the equivalence point. The concentration of Fe2+ in the original solution is…
(A) 0.0010 M
(B) 0.0056 M
(C) 0.028 M
(D) 0.090 M
(E) 0.14 M
Answer:
0.14 M
Explanation:
Volume of Fe2+ required for titration = 25 mL
Volume of MnO4- required for titration = 14 mL
Concentration of MnO4- = 0.05 M
We have to find the concentration of Fe2+.
Let it be x.
In a redox reaction:
Number of equivalence of oxidising agent = Number of equivalence of reducing agent.
Formula for number of equivalence: n*M*V
n = n-factor (number of electrons gained or lost by 1 mole of reducing or oxidising agent)
M = molarity
V = volume
n-factor of Fe2+ is 1 as it changes to Fe3+ by loosing 1 electron
n-factor of MnO4- is 5 as it changes to Mn2+ by gaining 5 electrons.
Number of equivalence of Fe2+ = Number of equivalence of MnO4-
1*(x)*25 = 5*0.05*14
x = 0.14 M
The concentration of Fe2+ in the original solution is 0.14 M. This is calculated using stoichiometry and the molarity of the MnO4− solution used in the reaction.
Explanation:The given question is about stoichiometry and molarity in a chemistry experiment involving a redox reaction of Fe2+ and MnO4−. To determine the concentration of Fe2+ in the original solution, we will use the reaction stoichiometry and the information given about the volume and concentration of the MnO4− solution. From the balanced equation, 5 moles of Fe2+ react with 1 mole of MnO4−. We know that the number of moles can be calculated as the product of volume (in litres) and molarity. Hence the moles of MnO4− is 14.0 mL (or 0.014 L) * 0.050 mol/L = 0.0007 mol. Therefore, the moles of Fe2+ = 5 * moles of MnO4− = 5 * 0.0007 mol = 0.0035 mol. The volume of Fe2+ solution is 25.0 mL or 0.025 L. Therefore, the concentration (molarity) of Fe2+ in the original solution = moles/volume = 0.0035 mol/0.025 L = 0.14 M. Therefore, the answer is (E) 0.14 M.
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Percent Ionization
Percent ionization for a weak acid (HA) is determined by the following formula:
Percent ionization=[HA] ionized[HA] initial×100%
For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization.
A certain weak acid, HA, has a Ka value of 7.6×10−7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution.
Express your answer as a percent using two significant figures.
%
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Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.
%
Answer:
(a) 0.26 % (b) 0.80 %
Explanation:
(a)
Given that:
[tex]K_{a}=7.6\times 10^{-7}[/tex]
Concentration = 0.10 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.10&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.10-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]
[tex]7.2\times 10^{-7}=\frac{x^2}{0.10-x}[/tex]
[tex]7.2\left(0.10-x\right)=10000000x^2[/tex]
Solving for x, we get:
x = 0.00026 M
Percentage ionization = [tex]\frac{0.00026}{0.10}\times 100=0.26 \%[/tex]
(b)
Concentration = 0.010 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.010&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.010-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]
[tex]7.2\times 10^{-7}=\frac{x^2}{0.010-x}[/tex]
[tex]7.2\left(0.010-x\right)=10000000x^2[/tex]
Solving for x, we get:
x = 0.00008 M
Percentage ionization = [tex]\frac{0.00008}{0.010}\times 100=0.80 \%[/tex]
I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down.
If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)
A. 200
B. 330
C. 400
D. 110
E. 88
Answer:
The correct answer is option B.
Explanation:
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]C_1[/tex] = specific heat of metal = [tex]900 J/kg^oC[/tex]
[tex]C_2[/tex] = specific heat of coffee= [tex]4000 J/kg^oC[/tex]
[tex]m_1[/tex] = mass of metal = x
[tex]m_2[/tex] = mass of coffee = 0.3 kg
[tex]T_f[/tex] = final temperature of aluminum metal= [tex]110^oC[/tex]
[tex]T_1[/tex] = initial temperature of aluminum metal = [tex]-10^oC[/tex]
[tex]T_2[/tex] = initial temperature of coffee= [tex]140^oC[/tex]
Now put all the given values in the above formula, we get
[tex]x\times 900 J/kg^oC\times (110-(-10))^oC=-(0.3 kg\times 4000 J/kg^oC\times (110-140)^oC[/tex]
[tex]x=0.333 kg[/tex]
Mass of aluminum cubes = 0.3333 kg = 333.3 g
If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:
[tex]=\frac{333.3 g}{1 g}=333.3\approx 330[/tex]
330 cubes of aluminum cubes will be required.
Scientists plan to place detection devices outside Earth's atmosphere to detect radiations from a star. Which of the following best explains the type of detection device that the scientists would use.a. A radar, because high frequency microwaves are limited to the region outside Earth's atmosphereb. Special cameras, because high frequency UV rays are emitted by objects at very high temperaturesc. Special cameras, because high energy UV rays are emitted by objects having very high massd. A radar, because high energy microwaves are produced from very hot objects
Answer: the correct option is A (A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere)
Explanation:
A radar can be used for detection of devices outside Earth's atmosphere to detect radiations from a star. A radar system consists of a transmitter producing electromagnetic waves in the radio or microwaves domain, a transmitting antenna, a receiving antenna (often the same antenna is used for transmitting and receiving) and a receiver and processor to determine properties of the object(s) example the stars
Answer:
A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere
Explanation:
I took the test and its correct
1.When HCl(aq) and NaOH(aq) are mixed in a beaker, the beaker feels warm to the touch. What is known about the enthalpy of this reaction?a. The reaction is exothermic.b. The reaction is endothermic.c. Heat is absorbed from the surroundings.d. \DeltaH is positive.Part B 2. Consider this combination reaction: 2Mg(s)+O2(g)→2MgO(s) ΔH=−1204 kJ What is the enthalpy for the decomposition of 1 mole of MgO(s) into Mg(s) and O2(g)? View Available Hint(s) A. -602 kJ/mol B. 602 kJ/mol C. -1204 kJ/mol D. 1204 kJ/mol Part C 3. The enthalpy for the formation of 1 mole of NH3(aq) is -80.29 kJ/mol. What is the enthapy for the formation of 3 moles of NH3(aq)? View Available Hint(s) The enthalpy for the formation of 1 mole of is -80.29 . What is the enthapy for the formation of 3 moles of ? A -83.29 kJ B. -518×103 kJ C. -240.87 kJ D. -26.76 kJ
Answer:
1. The correct answer is option a.
2. The correct answer is option B.
3. The correct answer is option C.
Explanation:
1. When acid reacts with base heat is generated along with formation of salt and water.
[tex]HCl+NaOH\rightarrow NaCl+H_2O,\Delta H=Negative[/tex]
Those reaction in which heat released as a product is called exothermic reaction.Exothermic reaction have negative value of enthalpy of reaction
2. [tex]2Mg(s)+O_2(g)→2MgO(s), \Delta H=-1204 kJ[/tex]
If we reveres the equation we will have the reaction in which MgO is getting decomposed into Mg and oxygen gas.
[tex]2MgO\rightarrow 2Mg+O_2(g),\Delta H=1204 kJ[/tex]
Divide the whole equation by 2.
[tex]MgO\rightarrow Mg+\frac{1}{2}O_2(g),\Delta H=602 kJ[/tex]
602 kJ is the enthalpy for the decomposition of 1 mole of MgO(s).
3.
The enthalpy for the formation of 1 mole of liquid ammonia = -80.29 kJ/mol
So, enthalpy of formation of 3 moles of liquid ammonia :
[tex]3 mol\times (-80.29 kJ/mol)=-240.87 kJ[/tex]
-240.87 kJ is the enthapy for the formation of 3 moles of liquid ammonia.
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.
The order of coefficients of the reactants and products in the balanced reaction is 1,3,6,1,3,3.
[tex]BrO_3^- (aq) + Sn^{2+} + H^+ \longrightarrow Br^- + Sn^{4+} + H_2O.[/tex]
What is the balanced chemical equation?The chemical equation in which the number of atoms of reactants and products is equal on either side of the equation is known as a balanced chemical equation.
According to the law of conservation of mass, the total mass on the reactant side should be equal to the total mass on the product side in a chemical equation
The given unbalanced chemical equation is as follows:
[tex]BrO_3^- (aq) + Sn^{2+} + H^+ \longrightarrow Br^- + Sn^{4+} + H_2O.[/tex]
To balance this equation, the coefficients are placed as shown below:
[tex]BrO_3^- (aq) + 3Sn^{2+} +6 H^+ \longrightarrow Br^- + 3Sn^{4+} + 3H_2O.[/tex]
Therefore, the order of the coefficient of reactants and products in the balanced equation is (1,3,6,1,3,3).
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Your question was incomplete, most probably the complete question was,
Consider the reaction given as
BrO₃⁻ (aq) + Sn²⁺ + __ → Br⁻ + Sn⁴⁺ + __
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the following pH
4.9
2.1
7.8
11.5
Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = [tex]10^{-4.8}[/tex]
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = [tex]10^{-7.6}[/tex]
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = [tex]10^{-1.9}[/tex]
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = [tex]10^{1.8}[/tex]
[In-]/[HIn] = 63.10
Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below.
When 0.0 25 mol of Na is added to 100.00 g of water, the temperature of the resulting solution rises from 25.00°C to 35.75°C.
If the specific heat of the solution is 4.18 J/(g · °C), calculate ΔH for the reaction, as written.
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) ΔH= ?
Answer: The enthalpy change of the reaction is -361.6 kJ
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sodium = 0.025 moles
Molar mass of sodium = 23 g/mol
Putting values in above equation, we get:
[tex]0.025mol=\frac{\text{Mass of sodium}}{23g/mol}\\\\\text{Mass of sodium}=(0.025mol\times 23g/mol)=0.575g[/tex]
We are given:
Mass of water = 100.00 g
Mass of sodium = 0.575 g
Mass of solution = 100.00 + 0.575 = 100.575 g
To calculate the amount of heat absorbed, we use the equation:
[tex]q=m\times C\times \Delta T[/tex]
where,
q = amount of heat absorbed = ?
m = mass of solution = 100.575 g
C = specific heat capacity of solution = 4.18 J/g°C
[tex]\Delta T[/tex] = change in temperature = [tex](T_2-T_1)=(35.75-25.00)=10.75^oC[/tex]
Putting all the values in above equation, we get:
[tex]q=100.575g\times 4.18J/g^oC\times 10.75^oC=4519.34J=4.52kJ[/tex]
When heat is absorbed by the solution, this means that heat is getting released by the reaction.
Sign convention of heat:
When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.
For the given chemical reaction:
[tex]2Na(s)+2H_2O(l)\rightarrow NaOH(aq.)+H_2(g)[/tex]
When 0.025 moles of sodium is reacted, the heat released by the reaction is 4.52 kJ
So, when 2 moles of sodium will react, the heat released by the reaction will be = [tex]\frac{4.52}{0.025}\times 2=361.6kJ[/tex]
Hence, the enthalpy change of the reaction is -361.6 kJ
To calculate ΔH for the reaction, we can use the equation ΔH = q / n, where q is the heat transferred and n is the number of moles involved in the reaction. Given the mass of the solution, specific heat, and temperature change, we can calculate the heat transferred and use it to find ΔH.
Explanation:The given chemical reaction is:
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
To calculate ΔH for the reaction, we can use the equation:
ΔH = q / n
Where:
ΔH is the enthalpy changeq is the heat transferredn is the number of moles involved in the reactionGiven that 0.025 mol of Na is added to 100.00 g of water and the temperature rise is 10.75°C, we can calculate the heat transferred:
q = m × c × ΔT
Where:
m is the mass of the solution in grams (100.00 g in this case)c is the specific heat of the solution (4.18 J/(g · °C) in this case)ΔT is the temperature change (35.75°C - 25.00°C = 10.75°C in this case)After calculating q, we can use it to find ΔH:
ΔH = q / n
Substituting the values, we get:
ΔH = q / n = (m × c × ΔT) / n
So, ΔH for the reaction is the calculated value.
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1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00V. 2. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, The number of kilowatt-hours of electricity required to produce of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is a. 59.6 b. 0.0168 c. 0.0596 d. 19.9 e. 39.7
*Answer:
Option A: 59.6
Explanation:
Step 1: Data given
Mass of aluminium = 4.00 kg
The applied emf = 5.00 V
watts = volts * amperes
Step 2: Calculate amperes
equivalent mass of aluminum = 27 / 3 = 9
mass of deposit = (equivalent mass x amperes x seconds) / 96500
4000 grams = (9* amperes * seconds) / 96500
amperes * seconds = 42888888.9
1 hour = 3600 seconds
amperes * hours = 42888888.9 / 3600 = 11913.6
amperes = 11913.6 / hours
Step 3: Calculate kilowatts
watts = 5 * 11913.6 / hours
watts = 59568 (per hour)
kilowatts = 59.6 (per hour)
The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V
Rank the following materials from highest BP to lowest BP: a. N2, NaF, NH3, Ni, HIb. Ag, H20, Ar, HBr, KC c. LiF, Fe, F2. HF, HCI
Explanation:
Metallic solids are defined as the solids in which atoms of metals are held together by metallic bonds. These bonds are much stronger than an ionic bond.
In metallic solids, the electrons are delocalized in nature. On the other hand, ionic solids have strong bonds due to the presence of opposite charges on its combining atoms.
Molecular solids are defined as the solids in which atoms are combined together through Vander waal forces. Hence, molecular solids have low boiling point.
(a) It is known that [tex]N_{2}[/tex] is a molecular solid, NaF is an ionic solid, in [tex]NH_{3}[/tex] there exists hydrogen bonding between the molecules, and Ni is a metallic solid. Hence, the given materials are placed according to highest B.P. to lowest B.P. as follows.
Ni > NaF > [tex]NH_{3}[/tex] > HI > [tex]N_{2}[/tex]
(b) Ag is a metallic solid, and KCl is an ionic solid. Oxygen atom is more electronegative than bromine atom. So, the boiling point of [tex]H_{2}O[/tex] is more than the boiling point of HBr. Whereas Ar is a noble gas and it has low boiling point.
Hence, the given materials are placed according to highest B.P. to lowest B.P. as follows.
Ag > KCl > [tex]H_{2}O[/tex] > HBr > Ar
(c) Fe is also a metallic solid and electronegativity of fluorine is more than the electronegativity of chlorine. Hence, HF has high boiling point than HCl. And, [tex]F_{2}[/tex] being a covalent compound has weak intermolecular forces. So, the boiling point of [tex]F_{2}[/tex] will be the least.
Therefore, the given materials are placed from highest B.P. to lowest B.P. as follows.
Fe > LiF > HF > HCl > [tex]F_{2}[/tex]
Final answer:
To rank the boiling points from highest to lowest, identify the type of bonding and intermolecular forces present in each substance: ionic, metallic, hydrogen bonding, dipole-dipole interactions, or London dispersion forces. Ionic and metallic bonds generally lead to higher boiling points than molecular compounds.
Explanation:
The student has asked a Chemistry question regarding the ranking of the boiling points (BP) of different substances. Boiling points can be compared using intermolecular forces: ionic bonds, hydrogen bonds, dipole-dipole interactions, and London dispersion forces. Ionic compounds like NaF typically have higher boiling points than covalent compounds due to the strong electrostatic force between ions.
Metallic bonds found in metals like Ni also exhibit high boiling points. Among covalent compounds, the presence of hydrogen bonding, as found in NH3, would lead to a higher boiling point than compounds that rely solely on dipole-dipole interactions (such as HF) or London dispersion forces (like N₂ and HBr).
Given this, we would rank the boiling points from highest to lowest as:
a. NaF > Ni > NH₃ > HI > N₂,
b. Ag > HBr > KC > H₂O > Ar
c. F₂ > Fe > LiF > HCI > HF
assuming the compounds are in pure form and standard atmospheric pressure is taken into account.
Which has not been suggested as a reasonably practical way to store large amounts of hydrogen in relatively small spaces for its use as a fuel?
A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.
B. Absorb hydrogen onto activated charcoal; then heat the mixture to release the hydrogen.
C. Store it in the form of ionic metal hydrides, such as LiH, which release hydrogen gas when they react with water.
D. Encapsulate hydrogen molecules in fullerene molecules (large, carbon-based molecules that can act like cages) that may be heated later to release the hydrogen.
Answer: A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.
Explanation:
Current Hydrogen storage methods fall into one of two technologies;
physical storage where compressed hydrogen gas is stored under pressure or as a liquid; and chemical storage, where the hydrogen is bonded with another material to form a hydride and released through a chemical reaction.Physical storage solutions are commonly used technologies but are problematic when looking at using hydrogen to fuel vehicles. Compressed hydrogen gas needs to be stored under high pressure and requires large and heavy tanks. Also, liquid hydrogen boils at -253°C (-423°F) so it needs to be stored cryogenically with heavy insulation and actually contains less hydrogen compared with the same volume of gasoline.
Chemical storage methods allow hydrogen to be stored at much lower pressures and offer high storage performance due to the strong binding of hydrogen and the high storage densities. They also occupy relatively smaller spaces than either compressed hydrogen gas or liquified hydrogen. A large number of chemical storage systems are under investigation, which involve hydrolysis reactions, hydrogenation/dehydrogenation reactions, ammonia borane and other boron hydrides, ammonia, and alane etc.
Other practical storage methods being researched that focuses on storing hydrogen as a lightweight, compact energy carrier for mobile applications include;
Metal hydrides e.g. LiHNanostructured metal hydrides Non-metal hydrides Carbohydrates Synthesized hydrocarbons Aluminum Liquid organic hydrogen carriers (LOHC) Encapsulation , e.t.c.Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241.8 kJ 2 ) X ( s ) + 2 Cl 2 ( g ) ⟶ XCl 4 ( s ) Δ H 2 = + 361.7 kJ 3 ) 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) ⟶ HCl ( g ) Δ H 3 = − 92.3 kJ 4 ) X ( s ) + O 2 ( g ) ⟶ XO 2 ( s ) Δ H 4 = − 607.9 kJ 5 ) H 2 O ( g ) ⟶ H 2 O ( l ) Δ H 5 = − 44.0 kJ what is the enthalpy, Δ H , for this reaction? XCl 4 ( s ) + 2 H 2 O ( l ) ⟶ XO 2 ( s ) + 4 HCl ( g )
Answer:
-767,2kJ
Explanation:
It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ
The sum of (4) - (2) produce:
6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ
(6) + 4×(3):
7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ
(7) - 2×(1):
8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ
(8) - 2×(5):
9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= -767,2kJ
I hope it helps!
To calculate the enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g), we can use Hess's law and the enthalpy values of the given reactions.
Explanation:The enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g) can be calculated using Hess's law and the enthalpy values of the given reactions.
To represent the desired reaction, we can combine reactions 2, 3, 4, and 5 as follows:
X (s) + 2 Cl2 (g) ⟶ XCl4 (s) [multiply this reaction by 2] 1/2 H2 (g) + 1/2 Cl2 (g) ⟶ HCl (g) [multiply this reaction by 4] X (s) + O2 (g) ⟶ XO2 (s) [multiply this reaction by 2] H2 (g) + 1/2 O2 (g) ⟶ H2O (l)By summing these equations, we get the desired equation:
XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g)
--
The enthalpy change for the reaction is ΔH = ΔH2 + ΔH3 + ΔH4 + ΔH5.
How many electrons are transferred in the following reaction?
(The reaction is unbalanced.)
Mg(s) + Al3+(aq) → Al(s) + Mg2+(aq)
Answer:
6
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]Mg(s)+Al^{3+}(aq.)\rightarrow Al(s)+Mg^{2+}(s)[/tex]
The half cell reactions for the above reaction follows:
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-[/tex]
Reduction half reaction: [tex]Al^{3+}+3e^-\rightarrow Al[/tex]
Magnesium is loosing 2 electrons to form the magnesium cation. Thus, it is getting oxidized. Aluminum anion is gaining 3 electrons to form Aluminum. Thus, it is getting reduced.
Thus, balancing the half-reactions as:-
Oxidation half reaction: [tex]3Mg\rightarrow 3Mg^{2+}+6e^-[/tex]
Reduction half reaction: [tex]2Al^{3+}+6e^-\rightarrow 2Al[/tex]
Thus, total number of electrons transferred = 6
Answer:
3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al
In this reaction 6 electrons are transferred
Explanation:
Step 1: The half reactions
Mg -2e- ⇆ Mg^2+
Al^3+ +3e- ⇆ Al
Step 2: Balance both equations
3*(Mg -2e- ⇆ Mg^2+)
2(Al^3+ +3e- ⇆ Al)
3Mg -6e- ⇆ 3Mg^2+
2Al^3+ +6e- ⇆ 2Al
Step 3: The netto reaction
3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al
In this reaction 6 electrons are transferred
A voltaic cell is constructed in which the anode is a Zn|Zn2+ half cell and the cathode is a Fe2+|Fe3+ half cell. The half-cell compartments are connected by a salt bridge.(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)The anode reaction is: ____+_____ yeilds _____ +______The cathode reaction is: _____+_______ yeilds_____ +________The net cell reaction is: _____+ _______yeilds _______+______In the external circuit, electrons migrate _______the Fe2+|Fe3+ electrode _______the Zn|Zn2+ electrode.In the salt bridge, anions migrate ________ the Fe2+|Fe3+ compartment ________ the Zn|Zn2+ compartment.
In a voltaic cell set up with Zn and Fe half cells, the anode reaction is Zn (s) → Zn2+ (aq) + 2e-, the cathode reaction is Fe3+ (aq) + e- → Fe2+ (aq), and the net cell reaction is Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq). Electrons move from the Zn electrode to the Fe electrode and anions in the salt bridge migrate from the Zn compartment to the Fe compartment.
Explanation:In a voltaic cell constructed in which the anode is a Zn|Zn2+ and the cathode is a Fe2+|Fe3+ half cell, the reactions would look something like this:
Anode reaction: Zn (s) → Zn2+ (aq) + 2e-
Cathode reaction: Fe3+ (aq) + e- → Fe2+ (aq)
Net cell reaction: Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq)
In the external circuit, electrons migrate from the Zn|Zn2+ electrode to the Fe2+|Fe3+ electrode. In the salt bridge, anions migrate from the Zn|Zn2+ compartment to the Fe2+|Fe3+ compartment.
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which of the following pieces of laboratory glassware should be used to most accurately measure out a 25.00 ml sample of a solution?
Ii. Out of a 25 mL graduated cylinder, 150 mL beaker, and a 50 mL Erlenmeyer flask which has the least uncertainty (most accurate volume)? Explain iii. If you had to accurately measure 20 mL of a liquid which of the three pieces of glassware would you use, 25 mL graduated cylinder, 150 mL beaker or a 50 mL? Explain
To most accurately measure out a 25.00 mL sample of a solution, a 25 mL graduated cylinder should be used. The same graduated cylinder can also be used to accurately measure 20 mL of a liquid.
Explanation:In order to most accurately measure out a 25.00 mL sample of a solution, a 25 mL graduated cylinder should be used. A graduated cylinder is designed with calibrated markings that allow for precise measurement of volume. It typically has a smaller uncertainty or error compared to a beaker or an Erlenmeyer flask.
If you had to accurately measure 20 mL of a liquid, you would still use the 25 mL graduated cylinder. Even though the cylinder has a larger capacity, it can still accurately measure smaller volumes. It is important to note that using a piece of glassware with a larger capacity than the required volume ensures that there is no spillage during the measurement process.
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The most accurate piece of laboratory glassware to measure out a 25.00 mL sample of a solution would be a 25 mL graduated cylinder. If you had to accurately measure 20 mL of a liquid, the best choice among the three options would still be the 25 mL graduated cylinder.
Explanation:The most accurate piece of laboratory glassware to measure out a 25.00 mL sample of a solution would be a 25 mL graduated cylinder. A graduated cylinder is designed to measure liquid volume with good accuracy. The markings on the cylinder allow for precise measurements, usually to the nearest 0.1 mL. By using the markings and reading the bottom of the meniscus, you can determine the volume of the solution.
If you had to accurately measure 20 mL of a liquid, the best choice among the three options would still be the 25 mL graduated cylinder. Although the graduated cylinder has a larger capacity than the required volume, it still provides accurate measurements within its range. It allows for better precision compared to a beaker or an Erlenmeyer flask.
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The target diol is synthesized in one step from 1-methylcyclopentene, but your lab partner exhausted the supply of that alkene. Fortunately, you have plenty of isomers (C6H10) on hand from which to synthesize 1-methylcyclopentene and, ultimately, the diol. Provide the missing reagents and organic structures needed to most efficiently produce the target product.
Answer:
The reagents are [tex]CH_{3}CH_{3}O^{-},OsO_{4},NaHSO_{3}and H_{2}O[/tex].
Explanation:
1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.
This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.
The chemical reaction is as follows.
With the help of isomers of C6H10, such as cyclohexene, it's possible to synthesize 1-methylcyclopentene by methylation and subsequent dehydrohalogenation. Once 1-methylcyclopentene is produced, the diol can be synthesized via a dihydroxylation process involving an oxidizing agent.
Explanation:The synthesis of the target diol from 1-methylcyclopentene can be achieved even if the starting alkene is not available by using isomers of C6H10. The first step involves synthesizing 1-methylcyclopentene from an appropriate isomer. For instance, cyclohexene, an isomer of C6H10, can be first transformed into 1-methylcyclohexene through an acid-catalyzed methylation reaction. To get 1-methylcyclopentene, a controlled elimination (dehydrohalogenation) step is necessary. This will produce the alkene with the double bond at the correct location. Once you have synthesized 1-methylcyclopentene, creating the diol is straightforward by applying the reaction conditions for dihydroxylation, which add a hydroxyl group to each carbon of the double bond.
Let's illustrate:
Start with cyclohexene (C6H10).Add HCl and CH3OH as reagents under acidic conditions to produce 1-methylcyclohexene.Heat this product with a strong base (e.g., KOH) to induce dehydrohalogenation and result in 1-methylcyclopentene.Finally, add an oxidizing agent (e.g., KMnO4) under acidic conditions to create the target diol.Learn more about Organic Chemistry Synthesis here:https://brainly.com/question/14917187
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Acetylene (C2H2) has a tendency to lose two protons (H1) and form the carbide ion (C2 22), which is present in a number of ionic compounds, such as CaC2 and MgC2. Describe the bonding scheme in the C2 22 ion in terms of molecular orbital theory. Compare the bond order in C2 22 with that in C2.
Answer:
The C22- ion is stable
Explanation:
The C22- ion is a stable ion having a bond order of three. It has a favourable stabilization energy of 6∆ compared to 4∆ in C2. In the carbide ion carbon firms three bonds rather than two in dicarbon, hence the formation of the carbide ion is preferred. The molecular orbital configuration of the carbide ion is shown in the image attached.
Final answer:
The carbide ion (C2 2-) is formed from acetylene (C2H2) by loss of two protons and has a bonding scheme where the triple bond of acetylene is reduced to a double bond in the ion, yielding a bond order of 2 for the carbide ion compared to 3 for acetylene, according to molecular orbital theory.
Explanation:
The carbide ion (C2 2-), formed when acetylene (C2H2) loses two protons, has an interesting bonding scheme described by molecular orbital theory. In acetylene, there is a triple bond between the two carbon atoms consisting of one sigma (σ) bond and two pi (π) bonds. This triple bond is the result of the overlapping of sp hybrid orbitals for the σ bond and the side-by-side overlap of unhybridized p orbitals for the π bonds. When the two protons are removed to form the carbide ion, two additional electrons are added to the system which occupy the antibonding π orbitals (π*). In molecular orbital theory, the bond order is equal to the difference between the number of bonding electrons and antibonding electrons divided by two. Thus, in the carbide ion, the triple bond of acetylene is reduced to a double bond due to the additional electrons in the antibonding orbitals, which gives a bond order of 2 in C2 2- as compared to a bond order of 3 in C2.
The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O 2 from the air is 2.67 × 10 –4 M at sea level and 25°C, what is the solubility of O 2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm?
Assume the temperature is 25°C, and that the mole fraction of O 2 in air is 0.209 at both 12,000 ft and at sea level. 1.75 × 10–4 M 2.67 × 10–4 M 3.66 × 10–5 M 4.06 × 10–4 M
Answer:
S = 1,75x10⁻⁴M
Explanation:
It is possible to answer this question using Henry's law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The formula is:
S = k×Pi
Where S is solubility, K is Henry's constant and Pi is partial pressure.
At sea level, Pi of O₂ is:
1atm×0,209 = 0,209 atm. Replacing:
2,67x10⁻⁴M = k×0,209atm
k = 1,28x10⁻³M/atm
At 12,000ft Pi of O₂ is:
0,657atm×0,209 = 0,137 atm. Replacing:
S = 1,28x10⁻³M/atm×0,137atm
S = 1,75x10⁻⁴M
I hope it helps!
An analytical chemist weighs out of an unknown diprotic acid into a volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with solution. When the titration reaches the equivalence point, the chemist finds he has added of solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.
The molar mass of an unknown diprotic acid is determined through a titration procedure. Using the volume and molarity of NaOH, we calculate the moles of NaOH, then divide by 2 to get the moles of diprotic acid. Dividing the mass of the acid by the moles gives us the molar mass.
Explanation:In this problem, the student is asked to calculate the molar mass of an unknown diprotic acid. The task requires the understanding of acids, stoichiometry, and titration concepts in analytical chemistry.
A diprotic acid is an acid that can donate two protons, or hydrogen ions, per molecule in solution. As the question doesn't provide the exact values, I'll explain conceptually. The molar mass of the unknown acid is calculated by using the volume of NaOH needed to reach the equivalence point and the concentration of NaOH. We'd start by calculating the moles of NaOH used (moles = volume x molarity), then because it's a diprotic acid, for every mole of acid, two moles of NaOH are needed, so we'd divide the moles of NaOH by 2 to get moles of unknown acid (moles acid = moles NaOH / 2). Lastly to find the molar mass, divide the mass of the acid used by the moles of acid calculated. So, the molar mass = mass of acid / moles of acid.
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The titration analysis provides a method to calculate the molar mass of an unknown acid using the volume of a known solution required to reach the equivalence point of the reaction. The process involves incrementally adding the known solution to the unknown, and using the reaction stoichiometry to find the unknown concentration. The molar mass is then calculated from this concentration, the initial volume of solution, and the molecular formula of the reaction products.
Explanation:Based on the information provided, we need further details to accurately calculate the molar mass of the unknown diprotic acid. However, we can understand the general process involved in calculating such values using titration data.
The process is generally as follows:
Titration analysis is conducted by incrementally adding a solution of known concentration (the titrant) to the unknown solution (analyte).The titrant and analyte undergo a chemical reaction of known stoichiometry. The volume of titrant required to reach the 'equivalence point' of the reaction - where both the titrant and analyte have been completely reacted - allows us to calculate the concentration of the unknown analyte solution. The molar mass of the unknown acid can then be calculated using the resulting concentration, the initial volume of solution, and the defined molecular formula and mass of the reaction products. Learn more about Titration here:https://brainly.com/question/31271061
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Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil. The radical formed by abstraction of one of the doubly allylic hydrogens is an allylic radical that has three resonance structures. Complete one of these resonance structures by dragging bonds and electrons to their appropriate positions.
Linoleic acid is a polyunsaturated fatty acid that forms an allylic radical with three resonance structures when its doubly allylic hydrogens are abstracted by radicals. These types of fats are liquid at room temperature and are nutritionally significant due to the body's inability to produce them. They carry numerous health benefits, including heart health and anti-cancer properties.
Explanation:
Linoleic acid is a polyunsaturated fatty acid found in various fats and oils. The hydrogen atoms in its doubly allylic positions are readily abstracted by radicals, leading to the oxidative degradation of the fat or oil. The resulting radical, called an allylic radical, has three resonance structures. Completing one resonance structure involves correctly positioning bonds and electrons.
Linoleic acid, like other polyunsaturated fats, can have more than one double bond in its structure. These fats are usually liquid at room temperature, like canola oil. When a fatty acid has a double bond in its structure, it is classified as unsaturated. Polyunsaturated fatty acids such as omega-3 fatty acids are nutritionally important because the human body cannot manufacture them. They must be obtained from the diet.
Fatty acids such as omega-3 fatty acids have a number of health benefits. They reduce the risk of heart attacks, reduce blood triglyceride levels, decrease blood pressure, and prevent thrombosis by inhibiting blood clotting. They also have anti-inflammatory properties and may help lower the risk of certain cancers.
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Write Lewis structural formulas for the following molecules:
NH3 SO2
CH3OH HNO2
N2 CH2O
Answer : The Lewis-dot structure of the following molecules are shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
(1) The given molecule is, [tex]NH_3[/tex]
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]NH_3[/tex] = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
(2) The given molecule is, [tex]SO_2[/tex]
As we know that sulfur and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]SO_2[/tex] = 6 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(3) The given molecule is, [tex]CH_3OH[/tex]
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]CH_3OH[/tex] = 4 + 4(1) + 6 = 14
According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.
(4) The given molecule is, [tex]HNO_2[/tex]
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]HNO_2[/tex] = 1 + 5 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(5) The given molecule is, [tex]N_2[/tex]
As we know that nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]N_2[/tex] = 2(5) = 10
According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.
(6) The given molecule is, [tex]CH_2O[/tex]
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]CH_2O[/tex] = 4 + 2(1) + 6 = 12
According to Lewis-dot structure, there are 8 number of bonding electrons and 4 number of non-bonding electrons.
For which of the following processes will \DeltaΔS be negative?PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)MgO(s) + CO2(g) = MgCO3(s)CO2(aq) = CO2(g)C5H12(l) + 8 O2(g) = 5 CO2(g) + 6 H2O(g)
Answer
a) [tex]PbCl_2(s)\rightarrow Pb^{2+}(aq)+2Cl^-(aq)[/tex]: [tex]\Delta S[/tex] = +ve
b) [tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]: [tex]\Delta S[/tex] = -ve
c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]: [tex]\Delta S[/tex]= +ve
d) [tex]C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex]: [tex]\Delta S[/tex] = +ve
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
a) [tex]PbCl_2(s)\rightarrow Pb^(2+)(aq)+2Cl^-(aq)[/tex]
As solid is moving to ions form , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
b)[tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]
As gaseous reactants are converted to solid products , randomness decreases and thus sign of [tex]\Delta S[/tex] is negative.
c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]
As liquid is changing to gas randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
d) [tex]C_5H_{12}(l)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)[/tex]
As 8 moles of gaseous reactants are converted to 11 moles of gaseous products , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
What is the half-life of A? What will the pressure, initially 32.1 kPa, at
(a) 10 s,
(b) 10 min after initiation of the reaction?
Answer:
a) 32.09 kPa
b) 32.09 kPa
Explanation:
Given data:
rate constant [tex]= 3.56\times 10^{-7} s^{-1}[/tex]
initial pressure is = 32.1 kPa
half life of A is calculated as
[tex]t_{1/2} = \frac{ln 2}{k}[/tex]
[tex]t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}[/tex]
[tex]t_{1/2} = = 1.956 \times 10^6 s[/tex]
for calculating pressure we have follwing expression
[tex]ln p = ln P_o - kt[/tex]
[tex]P =P_o e^{-kt}[/tex]
a) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa[/tex]
b) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa[/tex]
A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?
Answer:
a) 231.9 °C
b) 100% Sn
c) 327.5 °C
d) 100% Pb
Explanation:
This is a mixture of two solids with different fusion point:
[tex]Tf_{Pb}=327.5 C[/tex]
[tex]Tf_{Sn}=231.9 C[/tex]
Given that Sn has a lower fusion temperature it will start to melt first at that temperature.
So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.
The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.
At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.
Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass of helium in a helium filled tire. What is the mass difference between the two? Assume that the volume of the tire is 855ml that is filled with a total pressure of 125psi, and that the temperature is 25 degrees Celsius. Also, assume an average molar mass for air of 28.8g/mol
Answer: The mass difference between the two is 7.38 grams.
Explanation:
To calculate the number of moles, we use the equation given by ideal gas follows:
[tex]PV=nRT[/tex]
where,
P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)
V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)
T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles = ?
Putting values in above equation, we get:
[tex]8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For air:Moles of air = 0.297 moles
Average molar mass of air = 28.8 g/mol
Putting values in equation 1, we get:
[tex]0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g[/tex]
Mass of air, [tex]m_1[/tex] = 8.56 g
For helium gas:Moles of helium = 0.297 moles
Molar mass of helium = 4 g/mol
Putting values in equation 1, we get:
[tex]0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g[/tex]
Mass of helium, [tex]m_2[/tex] = 1.18 g
Calculating the mass difference between the two:
[tex]\Delta m=m_1-m_2[/tex]
[tex]\Delta m=(8.56-1.18)g=7.38g[/tex]
Hence, the mass difference between the two is 7.38 grams.
Final answer:
The mass of air in the tire calculated using the ideal gas law is 8.4752 grams, and the mass of helium is 1.1778 grams; the mass difference is 7.2974 grams, with helium being the lighter gas.
Explanation:
To calculate the mass of air and helium in a bike tire, we will use the ideal gas law which relates pressure (P), volume (V), temperature (T), and the number of moles of gas (n) using the equation PV = nRT where R is the gas constant. For this scenario, we are given that the volume (V) of the tire is 0.855 L (converted from 855 ml), the pressure (P) is 125 psi (which we must convert to atmospheres, since the gas constant uses atmospheres), and the temperature (T) is 25 degrees Celsius (which we must convert to Kelvin).
First, let's convert the pressure to atmospheres. Since 1 atm = 14.7 psi, the pressure in atmospheres is 125 psi / 14.7 psi/atm = 8.5034 atm. Next, let's convert the temperature to Kelvin: T = 25 °C + 273.15 = 298.15 K.
Using the ideal gas law equation PV = nRT, and solving for n (the number of moles), we get n = PV / RT. Substituting the known values for helium (R = 0.0821 L atm/mol K) and air (assuming average R = same as for helium, as there are no significant differences for the purposes of this problem):
For Air: n = (8.5034 atm) * (0.855 L) / (0.0821 L atm/mol K × 298.15 K) = 0.294 moles of air
For Helium: We use the same calculation as for air, since the volume, pressure, and temperature are the same.
Now, using the molar mass, we calculate the mass of each gas:
The mass of air = n (number of moles of air) × molar mass of air = 0.294 moles × 28.8g/mol = 8.4752 g
The molar mass of helium = 4.0026 g/mol, so the mass of helium = n (number of moles of helium) × molar mass of helium = 0.294 moles × 4.0026 g/mol = 1.1778 g
The mass difference between air-filled and helium-filled tires = mass of air - mass of helium = 8.4752 g - 1.1778 g = 7.2974 g. Thus, filling tires with helium instead of air would make them lighter by 7.2974 g.
What is the Molecule Geometry of a ABE3 molecule?
A. linear
B. trigonal bipyramidal
C. trigonal planar
D. tetrahedral
The Molecule Geometry of an ABE3 molecule, with three electron groups and no lone pairs, is trigonal planar, with groups arranged 120° apart in a plane.
Explanation:The Molecule Geometry of a molecule with the designation ABE3, where 'A' represents the central atom and 'E' represents electron groups (or bonding domains) surrounding the central atom, relates to how these groups are spatially distributed around the central atom. In an ABE3 molecule configuration, with three electron groups around the central atom and no lone pairs (indicated by the '3'), all three groups are arranged to be as far apart from one another as possible. This arrangement forms trigonal planar geometry, where the groups adopt the positions at the corners of an equilateral triangle, each 120° apart and all in the same plane.