Calculate the ph of a 1.60 mch3nh3cl solution. kb for methylamine, ch3nh2, is 3.4×10−4m. calculate the of a 1.60 solution. for methylamine, , is . 12.39 8.82 1.61 5.18
The base dissociation constant or Kb is a value used to measure the strength of a specific base in solution. To determine the percent ionization of the substance we make use of the Kb given. Methylamine or CH3NH2 when in solution would form ions:
CH3NH2 + H2O < = > CH3NH3+ + OH-
Kb is expressed as follows:
Kb = [OH-] [CH3NH3+] / [CH3NH2]
Where the terms represents the concentrations of the acid and the ions.
By the ICE table, we can calculate the equilibrium concentrations,
CH3NH2 CH3NH3+
OH-
I 1.60 0
0
C -x +x
+x
--------------------------------------------------
E 1.60-x x
x
Kb = [OH-] [CH3NH3+] / [CH3NH2] = 3.4×10−4
Solving for x,
x = [OH-] = 0.023 M
pH = 14 + log 0.023 = 12.36
Therefore, the first option is the closest one.
Answer:
5.2
Explanation:
Calculate the pH of a 1.60 M CH₃NH₃Cl solution. Kb for methylamine, CH₃NH₂, is 3.7 × 10⁻⁴.
CH₃NH₃Cl is a strong electrolyte that ionizes according to the following equation.
CH₃NH₃Cl(aq) → CH₃NH₃⁺(aq) + Cl⁻(aq)
The concentration of CH₃NH₃⁺ will also be 1.60 M (Ca). CH₃NH₃⁺ is the conjugate acid of CH₃NH₂. We can find its acid dissociation constant (Ka) using the following expression.
Ka × Kb = Kw
Ka × 3.7 × 10⁻⁴ = 1.0 × 10⁻¹⁴
Ka = 2.7 × 10⁻¹¹
The acid dissociation of CH₃NH₃⁺ can be represented through the following equation.
CH₃NH₃⁺(aq) ⇄ CH₃NH₂(aq) + H⁺(aq)
For a weak acid, we can find the concentration of H⁺ using the following expression.
[H⁺] = √(Ka × Ca) = √(2.7 × 10⁻¹¹ × 1.60) = 6.6 × 10⁻⁶ M
The pH is:
pH = -log [H⁺] = -log (6.6 × 10⁻⁶) = 5.2
How many grams are in 4.00 mol of sodium chloride (nacl)?
Mole measure the number of elementary entities of a given substance that are present in a given sample. The mass of 4.00 mole of sodium chloride is 233.76g.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance. One mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole of sodium chloride = mass ÷ Molar mass of sodium chloride
number of mole of sodium chloride=4mol
Molar mass sodium chloride =58.44 g/mole
Substituting the given values in the above equation, we get
4mole of sodium chloride=mass mass of sodium chloride÷ 58.44 g/mole
mass of sodium chloride = 4× 58.44 g/mole
mass of sodium chloride= 233.76g
Therefore the mass of 4.00 mole of sodium chloride is 233.76g.
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a blank is a possible explanation for a set of observations or answer to a scientific question
Given the formula c1v1=c2v2, where c indicates concentration and v indicates volume, which equation represents the correct way to find the concentration of the dilute solution (c2)
What property of liquid oxygen makes it especially difficult and potentially harmful to work with at home?
Write the formula for a complex formed between ag and nh3, with a coordination number of 2.
A complex formed between Ag (Silver) and NH3 (Ammonia) with a coordination number of 2 is [Ag(NH3)2]+
Explanation:In the world of chemistry, a complex compound is formed through the coordination of a central metal ion and surrounding ligands. Paying attention to your question, you're asking for the formula of a complex formed between Ag (Silver) and NH3 (Ammonia), with a coordination number of 2. The coordination number refers to the total number of ligand (NH3) links to a central ion (Ag). So, in this case, where there are two locations of the ammonia molecule bonded to the silver atom, the complex formula would be [Ag(NH3)2]+
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The formula for a complex formed between Ag (silver) and [tex]NH_3[/tex] (ammonia), with a coordination number of 2 for silver, would be [tex][Ag(NH_3)_2]^+[/tex].
To derive this formula, one must consider the charge of the silver ion and the nature of the ligand, [tex]NH_3[/tex]. Silver typically forms a 1+ ion, [tex]Ag^+[/tex]. Ammonia is a neutral ligand, meaning it does not carry a charge. When forming a complex, the ligands bond to the central metal ion, and the coordination number indicates how many ligands are bonded to the metal ion.
Given that silver has a coordination number of 2 in this complex, two NH3 molecules will bond to the [tex]Ag^+[/tex] ion. Since [tex]NH_3[/tex] is neutral, the overall charge of the complex will be the same as the charge of the silver ion, which is 1+. Therefore, the formula for the complex is [tex][Ag(NH_3)_2]^+[/tex].
Tetrahydrocannabinol (thc) is composed of 80.24% carbon, 9.62% hydrogen and 10.14% oxygen by mass. what is the empirical formula of thc?
For the reaction c2h6 (g) → c2h4 (g) + h2 (g) δh° is +137 kj/mol and δs° is +120 j/k ∙ mol. this reaction is ________. question 10 options: spontaneous only at high temperature spontaneous at all temperatures spontaneous only at low temperature nonspontaneous at all temperatures
Answer: spontaneous only at high temperature
Explanation: [tex]C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)[/tex]
The Gibbs free energy change is given by:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
when [tex]\DeltaG[/tex]= +ve, reaction is non spontaneous
[tex]\DeltaG[/tex]= -ve, reaction is spontaneous
[tex]\DeltaG[/tex]= 0, reaction is in equilibrium
[tex]\Delta H[/tex] = enthalpy change = endothermic = +137 KJ
[tex]\Delta S[/tex] = entropy change = +120 J/K
[tex]\Delta G=(+)-T(+)[/tex]
[tex]\Delta G=(+)(-ve)[/tex]
Now [tex]\Delta G= -ve[/tex] when [tex]T\Delta S[/tex] has more value than [tex]\Delta H[/tex]
Thus reaction is spontaneous at high temperatures.
When ΔH is positive and ΔS is positive, the reaction is spontaneous at high temperature.
In order to answer the question, we must define he following terms;
Exothermic reaction: This is a reaction in which the enthalpy change is negativeEndothermic reaction: This is a reaction in which he enthalpy change is positiveLet us recall that;
ΔG = ΔH - TΔS
Hence, when ΔH is positive and ΔS is positive, the reaction is spontaneous at high temperature.
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Bromobutane is treated with sodium methoxide in methanol at 323 k. draw the major elimination product of the reaction. make sure to consider the stereochemistry of the reaction.
The major elimination product of the reaction between bromobutane and sodium methoxide in methanol at 323 K can be obtained by removing the bromine atom and forming a double bond between the alpha and beta carbons. The stereochemistry of the reaction will depend on the starting configuration of the bromobutane molecule.
Explanation:The reaction between bromobutane and sodium methoxide in methanol at 323 K is an elimination reaction. It is likely to proceed via an E2 mechanism, which means that the elimination of a leaving group and the proton abstraction occur simultaneously.
In this reaction, the bromine atom will act as the leaving group and the methoxide ion will abstract a proton from the beta carbon.
The major elimination product of this reaction can be obtained by removing the bromine atom and forming a double bond between the alpha and beta carbons. The stereochemistry of the reaction will depend on the starting configuration of the bromobutane molecule.
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In the reaction of bromobutane with sodium methoxide, the major elimination product is 2-butene due to Zaitsev's rule favoring the formation of more substituted alkenes.
Explanation:When bromobutane is treated with sodium methoxide in methanol at 323 K, an elimination reaction (E2 mechanism) occurs to form an alkene. The major elimination product is the more substituted alkene due to Zaitsev's rule, which maximizes the number of alkyl groups attached to the double-bonded carbons. The reaction is as follows:
Bromobutane (CH3CH2CH2CH2Br) with Sodium Methoxide (NaOCH3) leads to:
(CH3)2C=CH2 - 2-butene (Major product because it is more substituted)CH3CH=CHCH3 - 1-butene (Minor product, less substituted)The major product will be 2-butene due to it having a higher degree of substitution, which is more stable and preferred in elimination reactions. The reaction involves the removal of beta hydrogen (from the 2nd carbon) and the bromine leaving group from the 1st carbon, resulting in the double bond between the 1st and 2nd carbons.
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Methane at 10 mpa and 300 k is heated at constant pressure until its volume has increased by 80 percent. determine the final temperature using the ideal gas equation of state and the compressibility factor. which of these two results is more accurat
To calculate the final temperature of methane gas heated at constant pressure with an 80% volume increase, one can use the ideal gas law or compressibility factor, with the latter expected to give a more accurate result.
Explanation:The student has asked a question that involves the heating of methane gas at a constant pressure and determining the final temperature using two different methods - the ideal gas law and the compressibility factor. Assuming an initial volume increase of 80 percent at constant pressure, the ideal gas law can be used by employing Charles's Law (V1/T1 = V2/T2), given that volume and temperature are directly proportional when pressure is held constant. To include real gas behavior, the Z factor or compressibility factor can be used, which corrects the ideal gas equation for the effect of intermolecular forces and the volume occupied by the gas molecules themselves.
For the ideal gas law, we can rearrange the equation to solve for the final temperature (T2).
Using compressibility factor, we would first need to determine the Z factor at the given conditions, and then apply this correction to the ideal gas equation to find the final temperature.
The use of the compressibility factor is expected to yield a more accurate result since it accounts for non-ideal behavior of gases, especially at high pressures and low temperatures where deviations from ideal gas behavior are more pronounced.
At the start of a measurement, a radioisotope has 10,000 unstable nuclei. over 2 days 5,000 of these unstable nuclei undergo radioactive decay to stable nuclei. what is the half life of this radioisotope?
A newly discovered element has two isotopes. one has an atomic weight of 120.9038 amu with 57.25% abundance. the other has an atomic weight of 122.8831 amu. what is the atomic weight of the element?
the atomic weight of the element : 122.22 amu
Further explanationThe elements in nature have several types of isotopes
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu
So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom
Mass atom X = mass isotope 1 . % + mass isotope 2.%
An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.
The 'amu' unit has now been replaced with a unit of 'u' only
for example, Carbon has 3 isotopes, namely ₆¹²C, ₆¹³C, and ₆¹⁴C
Element has two isotopes. 120.9038 amu with 57.25% abundance and the other has an atomic weight of 122.8831 amu
Mass atom element X = mass isotope 1 . % + mass isotope 2.%
Mass atom element X = 120.9038 . 57.25% + 122.8831. 42.75%
Mass atom element X = 69.22 + 52.53 = 122.22 amu
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Keywords: mass number, atomic mass, amu, isotope
. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?
Answer : The number of moles of ammonia gas is, 2.232 moles
Explanation : Given,
Volume of ammonia gas = 50 L
As we know that at STP,
1 mole of gas contains 22.4 L volume of gas.
As, 22.4 L volume of ammonia gas present in 1 mole of ammonia gas
So, 50 L volume of ammonia gas present in [tex]\frac{50}{22.4}=2.232[/tex] mole of ammonia gas
Therefore, the number of moles of ammonia gas is, 2.232 moles
The decomposition of dinitrogen pentoxide is described by the chemical equation 2 n2o5(g) → 4 no2(g) + o2(g) if the rate of disappearance of n2o5 is equal to 1.80 mol/min at a particular moment, what is the rate of appearance of no2 at that moment?
At that particular moment, the rate of appearance of NO₂ is 3.60 mol/min.
To determine the rate of appearance of nitrogen dioxide (NO₂) when dinitrogen pentoxide (N₂O₅) is decomposing, we can use the stoichiometry provided by the balanced chemical reaction:
[tex]2 \text{N}_2\text{O}_5 (g) \rightarrow 4 \text{NO}_2 (g) + \text{O}_2 (g)[/tex]
From this balanced equation, we observe the following molar ratios:
For every 2 moles of N₂O₅ that decompose, 4 moles of NO₂ are produced.
Therefore, the ratio of the change in concentration of N₂O₅ to the change in concentration of NO₂ is given by:
[tex]\frac{-\Delta [\text{N}_2\text{O}_5]}{2} = \frac{\Delta [\text{NO}_2]}{4}[/tex]
This can be rearranged to express the rate of formation of NO₂ based on the rate of disappearance of N₂O₅. Let's define the rate of disappearance of N₂O₅ as [tex]-\frac{d[\text{N}_2\text{O}_5]}{dt}[/tex].
Given:
Rate of disappearance of N₂O₅ = 1.80 mol/minAccording to the stoichiometry of the reaction, we have:
[tex]\frac{\Delta [\text{NO}_2]}{dt} \\= -\frac{4}{2} \times \frac{d[\text{N}_2\text{O}_5]}{dt} \\= 2 \times \frac{d[\text{N}_2\text{O}_5]}{dt}[/tex]
With the values plugged in:
[tex]\frac{d[\text{NO}_2]}{dt} = 2 \times (1.80 \text{ mol/min})[/tex]
[tex]\frac{d[\text{NO}_2]}{dt} = 3.60 \text{ mol/min}[/tex]
Which set of compounds would form a buffer in aqueous solution? drag each item to the appropriate bin. hints?
Compounds form a buffer in an aqueous solution when they consist of a weak acid and its conjugate base or a weak base and its conjugate acid. Some examples include acetic acid and sodium acetate or ammonia and ammonium chloride. The buffer's functionality is highest when the concentrations of these components are relatively equal and has the ability to resist changes in pH when minor quantities of acid or base are added.
In order for compounds to form a buffer in an aqueous solution, they must consist of a weak acid and its conjugate base or a weak base and its conjugate acid. One common example is a solutio of acetic acid (weak acid) and sodium acetate (conjugate base), represented as (CH3COOH + CH3COONa). Another example is a solution that includes ammonia (weak base) and ammonium chloride (conjugate acid), represented as (NH3(aq) + NH4Cl(aq)).
Buffers are most effective when their component concentrations are about equal. If the concentration of one component is less than about 10% of the other, the buffer's effectiveness diminishes. Buffer capacity, which refers to the amount of strong acid or base a buffer can neutralize, depends on the amount of acid and its conjugate base present in the solution.
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How much heat is needed to raise the temperature of a 5.0 gram block of aluminum from 22.0°c to 37.0°c?
Answer: 67.65 Joules
Explanation:
[tex]Q= m\times c\times \Delta T[/tex]
Q= heat gained
m= mass of the substance = 5.0 g
c = heat capacity of aluminium = 0.902 J/g ° C
[tex]\Delta T={\text{Change in temperature}}=(37-22)^oC=15^0C[/tex]
[tex]Q=5.0g\times 0.902J/g^oC\times 15^oC[/tex]
Q= 67.65 Joules
Thus heat gained by 5.0 g of aluminum is = 67.65 Joules
What is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2?
Given:
Mass of H2 = 5.22 kg = 5220 g
Mass of N2 = 31.5 kg = 31500 g
To determine:
Theoretical yield of NH3
Explanation:
The balanced chemical reaction is:
N2 + 3H2 → 2NH3
1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3
# moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles
# moles of H2 = 5200 g/ 1 g.mol-1 = 5200 moles
Therefore N2 is the limiting reagent
Based on the stoichiometry:
1 mole of N2 forms 2 moles of NH3
Thus, 1125 moles of N2 will yield : 1125 * 2 = 2250 moles of NH3
Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg
Ans: Theoretical yield of NH3 = 38.3 kg
Answer : The theoretical yield of ammonia is, 29.58 Kg
Solution : Given,
Mass of [tex]H_2[/tex] = 5.22 Kg = 5220 g
Mass of [tex]N_2[/tex]= 31.5 Kg = 31500 g
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex] .
Moles of [tex]H_2[/tex]= [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{5220g}{2g/mole}=2610moles[/tex]
Moles of [tex]N_2[/tex] = [tex]\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]
1125 moles of [tex]N_2[/tex] react with [tex]3\times 1125=3375[/tex] moles of [tex]H_2[/tex]
That means [tex]H_2[/tex] is a limiting reagent and [tex]N_2[/tex] is an excess reagent.
Now we have to calculate the moles of ammonia.
From the reaction we conclude that,
3 moles of [tex]H_2[/tex] react to give 2 moles of ammonia
2610 moles of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 2610=1740[/tex] moles of ammonia
Now we have to calculate the mass of ammonia.
[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]
[tex]\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg[/tex]
Therefore, the theoretical yield of ammonia is, 29.58 Kg
One mole of an ideal gas is sealed in a 22.4-l container at a pressure of 1 atm and a temperature of 273 k. the temperature is then increased to 311 k , but the container does not expand. what will the new pressure be? part a the most appropriate formula for solving this problem includes only which variables? enter the required variables, separated by commas (e.g., p,v,t).
The appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
Explanation:To solve this problem, we can use the combined gas law formula, which includes the variables pressure (P), volume (V), and temperature (T).
The combined gas law formula is as follows: P1V1/T1 = P2V2/T2
In this problem, the volume is constant, so we only need to consider pressure and temperature. Therefore, the appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
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1. After calculation, the new pressure of the gas is approximately 1.139 atm. 2.The variables involved are pressure (P) and temperature (T).
1. To determine the new pressure of the gas after the temperature increase while keeping the volume constant, we should use the combined gas law, which states:
P₁/T₁ = P₂/T₂
Given:
Initial pressure (P₁) = 1 atmInitial temperature (T₁) = 273 KFinal temperature (T₂) = 311 KWe need to find the final pressure (P₂)Rearrange the formula to solve for P₂:
[tex]\[P_2 = P_1 \cdot \frac{T_2}{T_1}\][/tex]
Substitute the given values:
[tex]P_2 = 1 \, \text{atm} \cdot \left( \frac{311 \, \text{K}}{273 \, \text{K}} \right)[/tex]
Calculate:
P₂ ≈ 1.139 atm
Therefore, the new pressure will be approximately 1.139 atm.
2. The most appropriate formula for solving this problem includes only the following variables: P, T.
Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.
0.275 m
11.7 m
0.990 m
0.857 m
Final answer:
To calculate the molality of the solution, divide the number of moles of lithium chloride (0.990 moles) by the mass of the water in kilograms (3.6 kg), resulting in a molality of 0.275 m.
Explanation:
The molality of a solution is calculated by finding the number of moles of solute and dividing by the mass of the solvent in kilograms. To find the molality of the solution with 42 grams of lithium chloride (LiCl) dissolved in 3.6 kg of water, we first convert grams to moles using the molar mass of LiCl.
The molar mass of LiCl is 42.39 g/mol. So, if we divide 42 grams by the molar mass, we get the number of moles of LiCl:
Moles of LiCl = 42 g / 42.39 g/mol = 0.990 moles of LiCl
We then use this number of moles and the mass of the solvent in kilograms to find the molality:
Molality (m) = Moles of solute / Mass of solvent (kg) = 0.990 moles / 3.6 kg = 0.275 m
Therefore, the molality of the solution is 0.275 m.
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 35.9°c.
The root mean square velocity (RMS velocity) is a method to find a single velocity value for the particles. The average velocity of gas particles in a mixture is found using the root mean square velocity formula:
μrms = sqrt (3RT/M)
Where,
R = universal gas constant = 8.3145 (kg·m^2/s^2)/K·mol
T = temperature = 35.9 ˚C = 309.05 K
M = molar mass = 28 * 10^-3 kg / mol
Substituting the given values into the equation:
μrms = sqrt [(3 * 8.3145 (kg·m^2/s^2) /K·mol) * (309.05 K) / (28 * 10^-3 kg / mol))]
μrms = sqrt (275,313.8813 m^2)
μrms = 524.7 m / s (ANSWER)
Which neutral atom is isoelectronic with o+?
The isoelectronic, neutral atom is Nitrogen (N), because both have the same number of electrons, which is 7.
To determine which neutral atom is isoelectronic with an [tex]O^+[/tex] ion, we first need to understand the term 'isoelectronic'. Electron counts are the same for isoelectronic atoms and ions.
Oxygen ([tex]O[/tex]) has an atomic number of 8, meaning a neutral oxygen atom has 8 electrons. When it forms an [tex]O^+[/tex] ion, it loses one electron, resulting in 7 electrons. We need to find a neutral atom that also has 7 electrons.Nitrogen (N) has an atomic number of 7, implying that a neutral nitrogen atom has 7 electrons. Hence, the neutral atom that is isoelectronic with [tex]O^+[/tex] is Nitrogen ([tex]N[/tex]).Why does potassium permanganate diffuse faster than methylene blue?
Diffusion of a substance is affected by its molecular weight and mobility of molecules. Potassium permanganate diffuse faster than methylene blue because, it is lighter than methylene blue.
What is diffusion ?Diffusion is the process of movement of molecules from from higher concentration area to that of lower concentration. The movement depends on the molecular weight, its density and mobility of its ions.
Potassium permanganate molecules are smaller and have lesser weight (150 g/mol) than methylene blue. Methylene blue is having higher molecular weight (319.85 g/mol) since it contains big aromatic rings.
Therefore, potassium permanganate diffuse faster than methylene blue.
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How much heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k?
528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
What is heat ?Heat is the result of the movement of kinetic energy within a material or an item, or from an energy source to a material or an object. Radiation, conduction, and convection are the three mechanisms through which such energy can be transferred.
Heat is the quantity of energy that is transferred from one structure to its surrounds as a result of a change in temperature. Heat is the transfer of kinetic energy between one media or item and another, or from an energy source to a medium or object.
The total amount of heat required
∑q = q1 + q2 + q3
q1 = mCΔT
= 475 g x 0.902 J/g-oC x ( 933.47 - 295 ) K
= 273,552.47J
= 273 kJ
ΔT = ( 933.47 - 295 ) K
= 638.47 K
ΔT = ( 933.47 K -273) - (295 K - 273)
= 638.47°C
For ΔT temperature in K is same as in °C
q2 = mol ΔHf
mol = 475 g x 1 mol /27g
= 17.6 mol
q2 = mol ΔHf = 17.6 mol x 10.79 kJ/mol
= 189.9 kJ
q3 = mCΔT
= 0.475 kg x 1.18 kJ/kg-oC (1050 - 933.47)K
= 65.3 kJ
∑q = q1 + q2 + q3
= 273 + 189.9 + 65.3
= 528.2 kJ
Thus,528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
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How many hydrogen atoms are in 5.20 mol of ammonium sulfide?
There are approximately 2.08 x 10^25 hydrogen atoms in 5.20 mol of ammonium sulfide.
Explanation:To determine the number of hydrogen atoms in 5.20 mol of ammonium sulfide (NH4)2S, we need to consider the molar ratio of hydrogen atoms to moles of ammonium sulfide. In the formula NH4)2S, there are 8 hydrogen atoms per 1 molecule. To find the number of hydrogen atoms, multiply the number of moles of ammonium sulfide by the Avogadro's number (6.022 x 1023 atoms/mol) and then multiply by the number of hydrogen atoms per molecule. So, the number of hydrogen atoms in 5.20 mol of ammonium sulfide is:
5.20 mol x 6.022 x 1023 atoms/mol x 8 atoms = 2.08 x 1025 hydrogen atoms.
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In [tex]5.20[/tex] moles of ammonium sulfide, there are approximately [tex]2.51 x 10^{25[/tex] hydrogen atoms.
To determine the number of hydrogen atoms in [tex]5.20[/tex] moles of ammonium sulfide [tex](\text{NH}_4)_2\text{S}[/tex], we must comprehend the compound's makeup.
First, note the formula of ammonium sulfide: [tex](\text{NH}_4)_2\text{S}[/tex].This indicates it contains two ammonium ions [tex]\text{(NH}_4^+)[/tex] and one sulfide ion ([tex]\text{S}^{2-}[/tex]).
Each ammonium ion ([tex]\text{NH}_4^+[/tex]) has 4 hydrogen atoms.Since there are two ammonium ions in each formula unit of ammonium sulfide, there are a total of 8 hydrogen atoms per formula unit of [tex](\text{NH}_4)_2\text{S}[/tex].
Now determine how many hydrogen atoms there are in total in [tex]5.20[/tex] moles of [tex](\text{NH}_4)_2\text{S}[/tex].Each mole of[tex](\text{NH}_4)_2\text{S}[/tex] contains Avogadro's number ([tex]6.022 \times 10^{23[/tex]) of formula units.
Thus, [tex]5.20[/tex] moles of [tex](\text{NH}_4)_2\text{S}[/tex] contain:
[tex](5.20 \, \text{moles}) \times (6.022 \times 10^{23} \, \text{formula units/mole}) = 3.13144 \times 10^{24} \, \text{formula units}[/tex]
In the reaction of 1.15 g of salicylic acid, molar mass 138.12 g/mol, with 2.83 g of acetic anhydride, molar mass 102.10 g/mol, a student obtained 1.21 g of acetylsalicylic acid, molar mass 180.17 g/mol. What is the percent yield?
The
balanced reaction would be written as:
C7H6O3 + C4H6O3--->C9H8O4 + HC2H3O2
To determine the percent yield, we need to first
determine the theoretical yield if the reaction were to proceed completely.
Then, we divide the actual yield that is given to the theoretical yield times
100 percent. The limiting reactant from the reaction would be salicylic acid.
We do as follows:
Theoretical yield: 1.15 g C7H6O3 ( 1 mol /
138.12 g ) ( 1 mol C9H8O4 / 1 mol C7H6O3 ) ( 180.17 g / mol ) = 1.50 g C9H8O4 should be produced
Percent yield = 1.21 / 1.50 x 100 = 80.66 %
Thus, only 80.66% of the
theoretical C9H8O4 is being produced.
The percent yield of the reaction is 14549%.
Explanation:To find the percent yield, we need to compare the actual yield (1.21 g of acetylsalicylic acid) to the theoretical yield, which is the maximum amount of acetylsalicylic acid that could be produced based on the reactants and their stoichiometry.
We can calculate the theoretical yield using the molar ratios of the reactants. The balanced equation for the reaction is:
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
From the equation, we can see that 1 mole of salicylic acid reacts with 1 mole of acetic anhydride to produce 1 mole of acetylsalicylic acid. Therefore, we can use the moles of salicylic acid and acetic anhydride to calculate the theoretical yield of acetylsalicylic acid.
The moles of salicylic acid can be calculated using its mass and molar mass:
moles of salicylic acid = mass of salicylic acid / molar mass of salicylic acid = 1.15 g / 138.12 g/mol = 0.00833 mol
The moles of acetic anhydride can be calculated using its mass and molar mass:
moles of acetic anhydride = mass of acetic anhydride / molar mass of acetic anhydride = 2.83 g / 102.10 g/mol = 0.0278 mol
Since the balanced equation shows a 1:1 molar ratio between salicylic acid and acetylsalicylic acid, the moles of salicylic acid and acetic anhydride are equal to the moles of acetylsalicylic acid produced.
Therefore, the theoretical yield of acetylsalicylic acid is 0.00833 mol.
The percent yield can be calculated using the formula:
percent yield = (actual yield / theoretical yield) * 100
Substituting the values, we get:
percent yield = (1.21 g / 0.00833 mol) * 100 = 14549%
The percent yield is 14549%.
Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.
The outer electron configurations for alkali metals, alkaline earth metals, halogens, and noble gases are described. Examples are given for each group of elements.
Explanation:The outer electron configurations for:
(a) Alkali metals: One s electron in the outermost shell. For example, the alkali metal with 11 protons and a mass number of 23 is potassium (K) with the electron configuration [Ar]4s1.(b) Alkaline earth metals: Two s electrons in the outermost shell. For example, the alkaline earth metal with 88 electrons and 138 neutrons is radium (Ra) with the electron configuration [Rn]7s2.(c) Halogens: One p electron in the outermost shell. For example, the isotope with 33 protons and 40 neutrons is arsenic (As) with the electron configuration [Ar]4s23d104p3.(d) Noble gases: Completely filled outermost shell. For example, the noble gas element with 75 neutrons in its nucleus and 54 electrons in the neutral atom is xenon (Xe) with the electron configuration [Kr]5s24d105p6.Learn more about outer electron configurations here:https://brainly.com/question/32534614
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Which of the following is the last step in performing a titration
1) finding out which pH indicatorworks
2) determining the concentration of an unknown base
3) finding the number of moles of a product produced in a reaction
4) determining the molecular masses of the products in the reaction
Answer: determining the concentration of an unknown base
Explanation: The last step in performing a titration will lead to the determination of the concentration of the unknown base or an unknown quantity.
The other steps like finding out which pH indicator works, finding the number of moles of a product produced in a reaction and determining the molecular masses of the products in the reaction comes in between the procedure.
The theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0g. what is the percent yield? 178 31.5 75.0 56.0 none of the above
Final answer:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. For the reaction with a theoretical yield of 75.0 grams and an actual yield of 42.0 grams, the percent yield is 56.0%.
Explanation:
To calculate the percent yield of a chemical reaction, you will need to divide the actual yield by the theoretical yield and then multiply the result by 100 to get a percentage. For the given reaction, if the theoretical yield is 75.0 grams and the actual yield is 42.0 grams, the percent yield can be calculated as follows:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Percent Yield = (42.0g / 75.0g) × 100
Percent Yield = 0.56 × 100
Percent Yield = 56.0%
Therefore, the correct answer from the options provided is 56.0%.
The world's oceans contain approximately 7.6 x 1019 moles of water molecules. Based on this, how many water molecules are there in the world's oceans?
Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5).
Final answer:
The pH of a 0.250 M solution of NaC2H3O2 is approximately 5.65.
Explanation:
The pH of a solution can be calculated using the expression: pH = -log[H⁺], where [H⁺] is the concentration of hydrogen ions in the solution. In this case, we have a 0.250 M solution of NaC2H3O2, which is the sodium salt of acetic acid (HC2H3O2). To calculate the pH, we need to first find the concentration of [H⁺].
Since NaC2H3O2 is a salt, it dissociates completely in water. The sodium ion (Na⁺) is a spectator ion and does not contribute to the pH. The C2H3O2⁻ ion reacts with water, forming HC2H3O2 and OH⁻ ions:
C2H3O2⁻ + H2O ⇌ HC2H3O2 + OH⁻
The equilibrium expression for the reaction is Ka = [HC2H3O2][OH⁻]/[C2H3O2⁻]. Given the value of Ka for HC2H3O2 as 1.80x10^-5, we can use this expression to find the concentrations of HC2H3O2 and OH⁻, and then determine the concentration of [H⁺]. Finally, we can calculate the pH using the formula pH = -log[H⁺].
After solving the equilibrium expression and performing the necessary calculations, we find the concentration of [H⁺] to be approximately 2.27x10^-6 M. Substituting this value into the pH formula, we get a pH of approximately 5.65.