Answer:
p(N2O4) = 0.318 atm
p(NO2) = 7.17 atm
Explanation:
Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?
Step 1: Data given
Kc = 0.619
Temperature = 45.0 °C
Mass of N2O4 = 50.0 grams
Volume = 2.10 L
Molar mass N2O4 = 92.01 g/mol
Step 2: The balanced equation
N2O4 ⇔ 2NO2
Step 3: Calculate moles N2O4
Moles N2O4 = 50.0 grams / 92.01 g/mol
Moles N2O4 = 0.543 moles
Step 4: The initial concentration
[N2O4] = 0.543 moles/2.10 L = 0.259 M
[NO2]= 0 M
Step 5: Calculate concentration at the equilibrium
For 1 mol N2O4 we'll have 2 moles NO2
[N2O4] = (0.259 -x)M
[NO2]= 2x
Step 6: Calculate Kc
Kc = 0.619= [NO2]² / [N2O4]
0.619 = (2x)² / (0.259-x)
0.619 = 4x² / (0.259 -x)
x = 0.1373
Step 7: Calculate concentrations
[N2O4] = (0.259 -x)M = 0.1217 M
[NO2]= 2x = 0.2746 M
Step 8: The moles
Moles = molarity * volume
Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles
Moles NO2 = 0.2746 M * 2.10 = 0.577 moles
Step 9: Calculate partial pressure
p*V = n*R*T
⇒ with p = the partial pressure
⇒ with V = the volume = 2.10 L
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature = 45 °C = 318 K
p = (nRT)/V
p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10
p(N2O4) = 0.318 atm
p(NO2) = (0.577 *0.08206 * 318)/ 2.10
p(NO2) = 7.17 atm
A mixture of water and graphite is heated to 600 K. When the system comes to equilibrium, it contains 0.13 mol of H2, 0.13 mol of CO, 0.43 mol of H2O, and some graphite. Some O2 is added to the system, and a spark is applied so that the H2 reacts com- pletely with the O2. Find the amount of CO in the flask when the system returns to equilibrium.
Explanation:
Expression to calculate the value of for the given reaction is as follows.
And, it is given that
[CO] = [tex][H_{2}][/tex] = 0.13 mol
= 0.43 mol
Putting the given values into the above formula as follows.
[tex]K_{c} = \frac{[CO][H_{2}]}{[H_{2}O]}[/tex]
= [tex]\frac{0.13 \times 0.13}{0.43}[/tex]
= 0.04
When additional amount of is added then all of has reacted.
So, new = 0 mols
new = 0.43 + 0.13 = 0.56 mols
The reaction equation is as follows.
[tex]C + H_{2}O \rightleftharpoons CO + H_{2}[/tex]
Initial: - 0.56 0.13 0
Change: - -x +x +x
Equilibm.: - 0.56 - x 0.13 + x x
So,
0.04 = [tex]\frac{(0.13 + x)(x)}{(0.56-x)}[/tex]
[tex]0.0224 - 0.04x = x^{2} + 0.13x[/tex]
[tex]x^{2} + 0.17x - 0.0224[/tex] = 0
x = 0.087 mols
Therefore, the amount of [CO] at equilibrium is as follows.
0.13 + 0.087
= 0.217 mols
thus, we can conclude that the amount of CO in the flask when the system returns to equilibrium is 0.217 moles.
Identify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues. a. an increase in blood acidity near the tissues b. an increase in blood temperature near the tissues c. the presence of a pressure gradient for oxygen d. an exchange of ions in the erythrocytes
Answer:
A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES
B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.
C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.
Explanation:
Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.
There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.
The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.
There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.
The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate·x-hydrate (CuSO4·xH2O), where x is an integer. Part of their practical exam is to determine this integer x. They are working in pairs, though Hermione is doing most of the work. This should be discouraged! After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4. What is the unknown integer x. Round the answer to the nearest integer.
Answer:
5
Explanation:
We can obtain the value of x by doing the following:
Mass of hydrated salt (CuSO4.xH2O) = 1.50g
Mass of anhydrous salt (CuSO4) = 0.96g
Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g
Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x
Mass of water(xH2O) molecules in the hydrate salt is given by:
xH2O/CuSO4.xH2O = 0.54/1.5
18x/(159.5 + 18x) = 0.36
Cross multiply to express in linear form
18x = 0.36 (159.5 + 18x)
18x = 57.42 + 6.48x
Collect like terms
18x — 6.48x = 57.42
11.52x = 57.42
Divide both side by 11.52
x = 57.42/11.52
x = 5
Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O
some plants grow in soils as high as 20% iron. If the iron is present in the form of Fe(OH)3, the plants can still be iron deficient. Explain how this can be true.
Explanation:
It is known that [tex]Fe(OH)_{3}[/tex] is insoluble in water. As a result, plants are not able to absorb [tex]Fe^{3+}[/tex] readily through osmosis.
Therefore, the [tex]Fe^{3+}[/tex] in [tex]Fe(OH)_{3}[/tex] would be released in acidic environments, using neutralization, Iron(III) ions can be released.
Hence, the easiest way is to add low concentrations of [tex]H_{2}SO_{4}[/tex] to the soil is as follows.
[tex]Fe(OH)_{3} + H_{2}SO_{4} \rightarrow H_{2}O + Fe_{2}(SO_{4})_{3}[/tex]
Thus, we can conclude that [tex]Fe_{2}(SO_{4})_{3}[/tex] is soluble and is good for plants too.
Plants can be iron deficient even when grown in soils with high iron content because the form of iron present, Fe(OH)3, is insoluble and not easily accessible by plants. Therefore, they may not be able to absorb enough iron to meet their requirements.
Explanation:The fact that plants can still be iron deficient even when grown in soils with high iron content can be explained by the form of iron present. In this case, the iron is in the form of Fe(OH)3. While plants require iron for proper growth and development, they can only absorb it in a specific form, called Fe²+. Fe(OH)3 is insoluble and cannot be easily accessed by plants. Therefore, even with high iron content, the plants may still not be able to absorb enough iron to meet their requirements, resulting in iron deficiency.
Learn more about Iron deficiency in plants here:https://brainly.com/question/36255477
#SPJ3
The gas phase reaction between NO2 and F2 is first order in [NO2] and first order in [F2l. What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?a.It would decrease by a factor of 2.b.It would increase by a factor of 4.c.It would increase by a factor of 2. d.It would remain unchanged. e.It would decrease by a factor of 4.
Answer:
e.It would decrease by a factor of 4.
Explanation:
first order in [NO2]
first order in [F2]
The rate is then given as;
Rate = k [NO2][F2]
where k = rate constant = And is constant for a reaction.
Let's insert some dummy values (Any values work, just be consistent);
[NO2] = 2
[F2] = 2
K = 3
Rate = 3*2*2
Rate = 12
What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?
[NO2] = 2 / 2 = 1
[F2] = 2 / 2 = 1
K = 3
Rate = 3*1*1
Rate = 3
Comparing both rates (12 and 3); the correct option is;
e.It would decrease by a factor of 4.
Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Complete the nuclear chemical equation below so that it describes this nuclear reaction.
____ → 30P15 + 1n0
Explanation:
An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.
Symbol of an alpha particle is [tex]^{4}_{2}\alpha[/tex]. Whereas a neutron is represented by a symbol [tex]^{1}_{0}n[/tex], that is, it has zero protons and only 1 neutron.
Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.
[tex]^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n[/tex]
In a nuclear reaction where an aluminum nuclide absorbs an alpha particle and emits a neutron to become a phosphorus nuclide, the balanced equation is ²⁷₁₃Al + ⁴₂α → ³⁰₁₅P + ¹₀n.
Explanation:When an aluminum nuclide absorbs an alpha particle and emits a neutron to transform into a phosphorus nuclide, we must consider both mass and atomic numbers for conservation in the nuclear reaction equation. Based on atomic numbers, aluminum with Z=13 will turn into phosphorus with Z=15 by absorbing an alpha particle (Helium nucleus) with a mass number of 4 and an atomic number of 2.
In terms of mass numbers, since a neutron (mass number of 1) is emitted from the aluminum, the mass number of the final phosphorus nuclide would be the mass number of aluminum (27) plus the mass number of the alpha particle (4), minus the mass number of the neutron (1), resulting in a phosphorus nuclide with a mass number of 30, which is phosphorus-30. The balanced nuclear equation is:
The elementary reaction 2 H 2 O ( g ) − ⇀ ↽ − 2 H 2 ( g ) + O 2 ( g ) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H 2 O , H2O, H 2 , H2, and O 2 O2 reach 0.0500 atm, 0.0500 atm, 0.00150 atm, 0.00150 atm, and 0.00150 atm, 0.00150 atm, respectively. What is the value of the equilibrium constant at this temperature?
Answer:
K = 0,00000135 = 1.35 * 10^-6
Explanation:
Step 1: Data given
The equilibrium constant, K, for any reaction is defined as the concentrations of the products raised by their coefficients divided by the concentrations of the reactants raised by their coefficients. In this case, the concentrations are given as partial pressures.
The partial pressures of H2O = 0.0500 atm
The partial pressures of H2 = 0.00150 atm
The partial pressures of O2 = 0.00150 atm
Step 2: The balanced equation
2H2O(g) ⇆ 2H2(g) + O2(g)
Step 3: Calculate K
K = [O2][H2]² / [H2O]²
K = 0.00150 * 0.00150² / 0.0500²
K = 0,00000135 = 1.35 * 10^-6
The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny sample of an artifact made of wood from an archeological dig is measured to be . The activity in a similar-sized sample of fresh wood is measured to be . Calculate the age of the artifact. Round your answer to significant digits.
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
Part A If 50.0 gg of N2O4N2O4 is introduced into an empty 2.12 LL container, what are the partial pressures of NO2NO2 and N2O4N2O4 after equilibrium has been achieved at 45∘C∘C?
The question is incomplete, here is the complete question:
At 45°C, Kc = 0.619 for the reaction N₂O₄(g) ⇌ 2 NO₂(g).
If 50.0 g of N₂O₄ is introduced into an empty 2.12 L container, what are the partial pressures of NO₂ and N₂O₄ after equilibrium has been achieved at 45°C?
Answer: The equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.
Explanation:
To calculate the number of moles, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = Pressure of the gas = ?
V = Volume of the gas = 2.12 L
w = Weight of the gas = 50.0 g
M = Molar mass of gas = 92 g/mol
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]45^oC=[45+273]K=318K[/tex]
Putting values in above equation, we get:
[tex]P\times 2.12L=\frac{50.0g}{92g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 318\\\\P=\frac{50.0\times 0.0821\times 318}{2.12\times 92}=6.693atm[/tex]
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure
[tex]K_c[/tex] = equilibrium constant in terms of concentration = 0.619
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]45^oC=[45+273]K=318K[/tex]
[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=(2-1)=1[/tex]
Putting values in above equation, we get:
[tex]K_p=0.619\times (0.0821\times 318)^{1}\\\\K_p=16.16[/tex]
For the given chemical equation:
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
Initial: 6.693
At eqllm: 6.693-x 2x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}[/tex]
Putting values in above expression, we get:
[tex]16.16=\frac{(2x)^2}{6.693-x}\\\\x=-7.59,3.56[/tex]
So, equilibrium partial pressure of [tex]NO_2=2x=(2\times 3.56)=7.12atm[/tex]
Equilibrium partial pressure of [tex]N_2O_4=(6.693-x)=(6.693-3.56)=3.133atm[/tex]
Hence, the equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.
2. (8 pts) Boric acid (H3BO3) has three hydrogens in a molecule, but effectively acts as a monoprotic acid (Ka = 5.8∙10-10), since the second and third Ka’s are negligible (less than 10-14). As any acid, H3BO3 will react with a strong base forming a salt (and possibly water). How many grams of boric acid and how many grams of NaOH are needed to prepare 1.00 L of a buffered solution with pH = 9.00 and a total concentration of boron 0.200 mol/L?
Answer:
7.85 g H₃BO₃
2.92 g NaOH
Explanation:
The strategy for solving this question is to first utilize the Henderson- Hasselbach equation to calculate the ratio of conjugate base concentration to weak acid:
pH = pKa + log [A⁻]/ [HA]
In this case:
pH = pKa + log [H₂BO₃⁻]/[H₃BO₃]
We know pH and indirectly pKa ( = - log Ka ).
9.00 = -log(5.8 x 10⁻¹⁰) + log [H₂BO₃⁻]/[H₃BO₃]
9.00 = 9.24 + log [H₂BO₃⁻]/[H₃BO₃]
log [H₂BO₃⁻]/[H₃BO₃] = - 0.24
taking inverse log function to both sides of the equation:
[H₂BO₃⁻]/[H₃BO₃] = 10^-0.24 = 0.58
We are also told we want to have a total concentration of boron of 0.200 mol/L, and if we call x the concentration of H₂BO₃⁻ and y the concentration of H₃BO₃, it follows that:
x + y = 0.200 ( since we have 1 Boron atom per formula of each compound)
and from the Henderson Hasselbach calculation, we have that
x / y = 0.58
So we have a system of 2 equations with two unknowns, which when solved give us that
x = 0.073 and y = 0.127
Because we are told the volume is one liter it follows that the number of moles of boric acid and the salt are the same numbers 0.073 and 0.127
gram boric acid = 0.127 mol x molar mass HBO₃ = 0127 mol x 61.83 g/mol
= 7.85 g boric acid
grams NaOH = 0.073 mol x molar mass NaOH = 0.073 x 40 g/mol
= 2.92 g NaOH
Ammonium phosphate ((NH4)3PO4) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H3PO4) with ammonia (NH3).
What mass of ammonium phosphate is produced by the reaction of 4.9 g phosphoric acid?
The mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid is 7.45 g.
Explanation:The question asks for the mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid. To determine the mass of ammonium phosphate produced, we need to balance the chemical equation and calculate the molar mass of both reactants and products.
The balanced equation for the reaction is:3H3PO4 + (NH4)OH → (NH4)3PO4 + 3H2O
The molar mass of phosphoric acid (H3PO4) is 97.99 g/mol. The molar mass of ammonium phosphate ((NH4)3PO4) is 149.0 g/mol.
Using the molar mass of phosphoric acid and the ratio of the reactants and products in the balanced equation, we can calculate the mass of ammonium phosphate produced.First, calculate the moles of phosphoric acid:
moles of H3PO4 = mass (g) / molar mass (g/mol)
moles of H3PO4 = 4.9 g / 97.99 g/mol = 0.050 moles
Since the stoichiometry of the reaction is 1:1 between phosphoric acid and ammonium phosphate, the moles of ammonium phosphate produced is also 0.050 moles.
Finally, calculate the mass of ammonium phosphate:mass of (NH4)3PO4 = moles of (NH4)3PO4 × molar mass of (NH4)3PO4
mass of (NH4)3PO4 = 0.050 moles × 149.0 g/mol = 7.45 g
Therefore, 7.45 g of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid.
Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)
(a) O2 (g)---->2/3O3(g) Kc=5.77x10^-9
(b) 3O2 (g)----->2O3(g) Kc=?
Answer: The value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]
Explanation:
The given chemical equation follows:
[tex]O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation is [tex]5.77\times 10^{-9}[/tex]
We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:
[tex]3O_2(g)\rightarrow 2O_3(g)[/tex]
The equilibrium constant for this reaction will be the cube of the initial reaction.
If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}[/tex]
Hence, the value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]
The equilibrium constant for equation (b) is 1.73x10^8.
Explanation:To calculate the equilibrium constant (Kc) for equation (b), we can use the relationship between the equilibrium constants of related reactions. The equation (b) is the reverse of equation (a), so the equilibrium constant for equation (b) is the reciprocal of the equilibrium constant for equation (a).
Therefore, the equilibrium constant (Kc) for equation (b) is 1 / (5.77x10^-9) = 1.73x10^8.
Learn more about Equilibrium constant here:https://brainly.com/question/32442283
#SPJ11
Given the equation: rate = k[H2O2]m[I-]n Which of the following statements is (are) true about the orders, m and n? A. m and n are independent from the molar coefficients of the reactants in the balanced chemical equation. B. m and n must be determined by experiment C. m and n are equal to each other in all cases. D. m and n are equal to the molar coefficients of H2O2 and I- in the balanced chemical reaction, respectively.
Answer:
A. m and n are independent from the molar coefficients of the reactants in the balanced chemical equation.
B. m and n must be determined by experiment.
Explanation:
rate = k[H2O2]^m × [I-]^n
The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant.
Either the differential rate law or the integrated rate law can be used to determine the reaction order of reactants from experimental data.
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
The question in incomplete, complete question is;
Determine the theoretical yield:
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
Answer:
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
[tex]Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)[/tex]
Moles of copper(II) nitrate = [tex]\frac{469 g}{187.5 g/mol}=2.5013 mol[/tex]
Moles of sodium sulfide = [tex]\frac{156 g}{78 g/mol}=2 mol[/tex]
According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.
Then 2 moles of sodium sulfide will react with:
[tex]\frac{1}{1}\times 2mol= 2 mol[/tex] of copper (II) nitrate
As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.
According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:
[tex]\frac{2}{1}\times 2mol=4 mol[/tex] sodium nitrate
Mass of 4 moles of sodium nitrate :
85 g/mol × 4 mol = 340 g
Theoretical yield of sodium nitrate = 340 g
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
Below is an attachment containing the solution.
Some people believe that if by placing a cold spoon in a cup of hot coffee, it will cool it enough to drink it comfortably. Let’s test this. If you have a silver spoon that has been chilled at T= 10.0 °C (let’s say mass = 100.0 g, assume 100% Ag, specific heat =0.235 J/g °C), and you place it in a 240. mL cup of coffee that is at T= 90.0 °C (a typical temperature at a McDonald’s restaurant).
What will the final temperature of the coffee? (Food for thought, if you spill this coffee, will it be hot enough to give you 3rd-degree burns?)
Answer:
[tex]\large \boxed{\text{88.1 $^{\circ}$C}}[/tex]
Explanation:
There are two heat transfers involved: the heat gained by the spoon and the heat lost by the coffee.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the spoon be Component 1 and the coffee be Component 2.
Data:
For the spoon:
[tex]m_{1} =\text{100.0 g; }T_{i} = 10.0 ^{\circ}\text{C; }\\C_{1} = 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the coffee:
[tex]m_{2} =\text{240.0 g; }T_{i} = 90.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
1. The relative temperature changes
[tex]\begin{array}{rcl}\text{Heat gained by spoon + heat lost by coffee} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{100.0 g}\times 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{240. g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\23.50\Delta T_{1} + 1004\Delta T_{2} & = & 0\\\end{array}[/tex]
[tex]\begin{array}{rcl}1004\Delta T_{2} & = & -23.50\Delta T_{1}\\\Delta T_{2} & = & -0.02340\Delta T_{1}\\\end{array}\\\text{(The temperature change for the coffee is about 1/40 that of the spoon.)}[/tex]
2. Final temperature of coffee
[tex]\Delta T_{1} = T_{\text{f}} - 10.0 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 90.0 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}\Delta T_{2} & = & -0.02340\Delta T_{1}\\T_{\text{f}} - 90.0 \, ^{\circ}\text{C} & = & -0.02340 (T_{\text{f}} - 10.0 \, ^{\circ}\text{C})\\& = & -0.02340T_{\text{f}} + 0.2340 \, ^{\circ}\text{C}\\T_{\text{f}} & = & -0.02340T_{\text{f}}+ 90.23 \, ^{\circ}\text{C}\\1.02430T_{\text{f}}& = & 90.23 \, ^{\circ}\text{C}\\\end{array}\\[/tex]
[tex]\begin{array}{rcl}T_{\text{f}} & = & \dfrac{ 90.23 \, ^{\circ}\text{C}}{1.02430}\\\\ & = & \mathbf{88.1 \,^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the coffee is $\large \boxed{\textbf{88.1 $\,^{\circ}$C}}$}\\\text{This is hot enough to cause third-degree burns in less than 1 s.}[/tex]
Using the principles of physics and the formula for heat transfer, the final temperature after introducing a cold spoon to a cup of hot coffee ends up being around 89.6°C. It's further noted that spilling coffee at this temperature could potentially cause 3rd-degree burns, although the specific outcome may vary.
Explanation:This question can be answered through the principles of energy conservation. In this situation, the chilled spoon will absorb heat from the hot coffee until they both reach a thermal equilibrium. This means they will eventually have the same temperature. The formula for this heat transfer can be expressed as Q (heat energy transferred) = mcΔT, where 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.
Applying this to the spoon and coffee respectively, we will end up with an equation: (mass of coffee * specific heat of coffee * (initial coffee temp - final temp)) = -(mass of spoon * specific heat of spoon * (final temp - initial spoon temp)). Considering the specific heat of water (essentially coffee without impurities) is approximately 4.186 J/g °C, the initial math leads us to a final temperature of around 89.6°C.
Mentioning the last thought in your question, it's likely that spilling coffee at this temperature could potentially cause 3rd-degree burns. The exact effect can vary depending on various factors such as exposure duration and skin sensitivity, but typically, exposure to liquid at a temperature above 70°C can cause severe burns in a matter of seconds.
Learn more about Heat Transfer here:https://brainly.com/question/13433948
#SPJ11
The radius of a iridium atom is 135 pm. How many iridium atoms would have to be laid side by side to span a distance of 2.33 mm? g
Answer:
Number of atoms, N = 17,259,259
Explanation:
Given:
Radius of Iridium = 135 pm
distance = 2.33 mm
To determine number of iridium atom that would be laid side by side to span a distance of 2.33 mm, we say let the number of the atom = N
[tex]N =\frac{Distance}{Iridium. Radius} = \frac{2.33X10^{-3}}{135 X10^{-12}} \\\\N = 17259259.26[/tex]
Therefore, If a iridium atom has a radius of 135 pm, then 17,259,259 atoms of Iridium would be laid side by side to span a distance of 2.33 mm.
Number of atoms, N = 17,259,259
17,259,259 atoms of iridium are present in a distance of 2.33 mm
First, to calculate the amount of iridium atoms in a distance of 2.33 mm it is necessary to divide the two values:
[tex]N = \frac{Distance}{Radius} [/tex]
As the element radius value is in pm, it is necessary to transform this unit to suit the other:
[tex]135pm = 135\times 10^{-9} mm[/tex]
Now, we can apply the values in the expression:
[tex]N = \frac{2.33}{135\times10^{-9}}[/tex]
[tex]N = 17,259,259 [/tex] atoms
So, 17,259,259 atoms of iridium are present in a distance of 2.33 mm.
Learn more about calculation of atoms in: brainly.com/question/3320707
The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0-ppb solution. What assumption did you have to make in your calculation? (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m3 of water?
Answer:
0.54 g of Pb in 60 m³ of [Pb] = 9 ppb
Explanation:
We need to know, that ppb is a sort of concentration that indicates (in value of volume) ng / mL (ppb = parts per billion)
9 ppb means that 9 ng are contained in 1mL of solution
We know that 1mL = 1 cm³
Let's convert 60 m³ to cm³ → 60 m³ . 1×10⁶ cm³ / 1m³ = 6×10⁷ cm³
Then, we can make a rule of three:
1 cm³ has 9 ng of Pb
Therefore in 6×10⁷ cm³ we must have ( 6×10⁷ cm³ . 9ng) / 1 cm³ =
5.4×10⁸ ng
We convert ng to g → 5.4×10⁸ ng . 1 g / 1×10⁻⁹ ng = 0.54 g
Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.
The maximum allowable concentration of lead in drinking water is 9.0 ppb, that is, 9.0 μg of Pb per liter of water.
We want to calculate the molarity (M) of the solution. We will use the following expression, assuming that the volume of water is equal to the volume of the solution.
M = mass Pb / molar mass Pb × liters of solution
M = (9.0 × 10⁻⁶ g) / (207.2 g/mol) × 1 L = 4.3 × 10⁻⁸ M
To calculate the mass of Pb in a 60 m³ pool, we need to consider that 1 m³ = 1000 L.
60 m³ Water × (1000 L Water/ 1 m³ Water) = 60,000 L Water
60,000 L Water × (9.0 × 10⁻⁶ g Pb/1 L Water) = 0.54 g Pb
Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.
Learn more about parts per billion here: https://brainly.com/question/13290253
A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The end point was obtained at about 24.8 mL. The concentration of the monoprotic acid is about ........ mol/L.
A) 25.0
B) 0.0600
C) 0.240
D) 0.120
E) None of the abov
Answer:
The concentration of the monoprotic acid is about 0.114 mol/L.
The correct answer is E none of the above
Explanation:
Step 1: Data given
Volume of a monoprotic acid = 25.0 mL = 0.025 L
Molarity of the NaOH solution = 0.115 M
The end point was obtained at about 24.8 mL.
Step 2: Calculate the concentration of the monoprotic acid
b*Ca*Va = a*Cb*Vb
⇒with B = the coefficient of NaOH = 1
⇒with Ca = the concentration of the monoprotic acid = ?
⇒with Va = the volume of the monoprotic acid = 0.025 L
⇒with a = the coefficient of the monoprotic acid = 1
⇒with Cb = the concentration of NaOH = 0.115M
⇒with Vb = the volume of NaOH= 0.0248 L
1*Ca*0.025 = 1*0.115*0.0248
Ca = (0.115*0.0248)/0.025
Ca = 0.114 M
The concentration of the monoprotic acid is about 0.114 mol/L.
The correct answer is E none of the above
An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically.
The reaction of compound A forming compound B was studied and the following data were collected:
Time (s) [A](M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016
a.) What is the average reaction rate between 0. and 1500. s?
b.) What is the average reaction rate between 200. s and 1200. s?
c.) What is the instantaneous rate of the reaction at t=800 s?
Answer:
a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.
b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.
c) Instantaneous rate of the reaction at t=800 s :
Instantaneous rate : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]
Explanation:
Average rate of the reaction is given as;
[tex]R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}[/tex]
a.) The average reaction rate between 0.0 s and 1500.0 s:
At 0.0 seconds the concentration was = [tex]A_1=0.184 M[/tex]
[tex]t_1=0.0s[/tex]
At 1500.0 seconds the concentration was = [tex]A_2=0.016 M[/tex]
[tex]t_2=1500 s[/tex]
[tex]R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s[/tex]
0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.
b.) The average reaction rate between 200.0 s and 1200.0 s:
At 0.0 seconds the concentration was = [tex]A_1=0.129 M[/tex]
[tex]t_1 =200.0 s[/tex]
At 1500.0 seconds the concentration was = [tex]A_2=0.019M[/tex]
[tex]t_2=1200 s[/tex]
[tex]R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s[/tex]
0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.
c.) Instantaneous rate of the reaction at t=800 s :
At 800 seconds the concentration was = [tex]A=0.031 M[/tex]
[tex]t =800.0 s[/tex]
Instantaneous rate : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]
Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed leave it blank. If no reaction occurs leave all boxes blank and click on "Submit". Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrochloric acid and potassium hydroxide are combined. (Use H+ instead of H3O+.)
Answer:
H+ + OH− --> H2O
Explanation:
Hydrochloric acid is represented by the chemical formular; HCl. This is an ionic substance so in water it breaks apart into hydrohrn ions; H+ and chloride ions; Cl−. It is a strong acid, hence it completely dissociates.
Potassium Hydroxide is also an ionic substance it also breaks apart in water into potassium ions; K+ and hydroxide ions; OH−. It is a strong base, hence it completely dissociates.
The complete ionic equation for the reaction is given as;
H+ + Cl− + K+ + OH− --> K+ + Cl− + H2O
The Hydrogen ion and the Hydroxide ions combine to form water.
The net ionic equation is given as;
H+ + OH− --> H2O
Cl- and K+ ions were cancelled out because they do not undergo any changes therefore are not part of the net ionic equation. They are referred to as spectator ions.
In the Millikan oil droplet experiment, the oil is sprayed from an atomizer into a chamber.The droplets are allowed to pass through the hole into the chamber so that their fall can be observed. The top and bottom of the chamber consist of electrically charged plates. The upper plate is positively charged, and the lower plate is negatively charged. X rays are introduced into the chamber so that when they strike the oil droplets, the droplets will acquire one or more negative charges. The electric field (voltage) is applied to the metal plates.Which of the following applies? Can have more than one answer.a. In the presence of an electric field, the negatively charged oil droplet moves freely toward the negatively charged plate.b. In the absence of an electric field, the oil droplet falls freely due to the gravitational force.c. If Fe is greater than Fg , the negatively charged oil droplet will move freely toward the negatively charged plate.d. If Fe is increased until it is equal to Fg , the negatively charged oil droplet will remain stationary.
Answer:
is this just a passage?
Explanation:
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a flask with of ethylene gas and of water vapor. When the mixture has come to equilibrium she determines that it contains of ethylene gas and of water vapor. The engineer then adds another of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to significant digits.
The question is incomplete, here is the complete question:
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.
A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.
The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
Explanation:
We are given:
Initial partial pressure of ethylene gas = 1.8 atm
Initial partial pressure of water vapor = 4.7 atm
Equilibrium partial pressure of ethylene gas = 1.16 atm
Equilibrium partial pressure of water vapor = 4.06 atm
The chemical equation for the reaction of ethylene gas and water vapor follows:
[tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]
Initial: 1.8 4.7
At eqllm: 1.8-x 4.7-x
Evaluating the value of 'x'
[tex]\Rightarrow (1.8-x)=1.16\\\\x=0.64[/tex]
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}[/tex]
[tex]p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm[/tex]
Putting values in above expression, we get:
[tex]K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136[/tex]
When more ethylene is added, the equilibrium gets re-established.
Partial pressure of ethylene added = 1.2 atm
[tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]
Initial: 2.36 4.06 0.64
At eqllm: 2.36-x 4.06-x 0.64+x
Putting value in the equilibrium constant expression, we get:
[tex]0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41[/tex]
Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.
So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm
Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
What is the concentration (M) of HCl in a solution that is prepared by dissolving 25.5 g of HCl in of water? (Assume volume does not change after adding HCl.)
Answer:
[HCl] = 0.035 M
Explanation:
COMPLETE QUESTION:
What is the concentration (M) of HCl in a solution that is prepared by dissolving 25.5 g of HCl in 20.0 L of water? (Assume volume does not change after adding HCl.)
Molarity → moles of solute in 1L of solution
As we assume volume does not change after adding HCl, solution's volume is 20L
We convert solute's mass to moles → 25.5 g. 1mol / 36.45 g = 0.699 moles
Molarity (mol/L) → 0.699 mol/20L = 0.035 M
To find the concentration of HCl, divide the mass of HCl (25.5 g) by its molar mass to get moles, then divide by the volume of the solution in liters, assuming a volume of 1L for simplicity.
Explanation:Firstly, the molar mass of HCl must be determined, which is approximately 36.46 g/mol. To find the concentration (Molarity, M) of HCl in the solution, the mass of HCl (25.5 g) must be converted into moles by dividing by the molar mass (moles of HCl = 25.5 g / 36.46 g/mol = 0.699 moles).
Then, the volume of water needs to be converted into liters (assuming it to be 1L for this example, as the exact volume was not provided). Finally, the concentration can be calculated by dividing the moles of HCl by the volume of the solution in liters (Concentration, M = 0.699 moles / 1L = 0.699 M)
2. High temperatures in the automobile engine cause nitrogen and oxygen gases from the air to combine to form nitrogen oxides (NO and NO2). What two acids in acid rain result from the nitrogen oxides in automobile exhaust
Answer:
1. HNO3
2. HNO2
Explanation:
Nitrogen acid may refer to any of:
1. Nitric acid, HNO3 or
2. Nitrous acid, HNO2
Acid rain is caused by a chemical reaction that begins when compound like nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain. nitrogen oxides dissolve very easily in water and can be carried very far by the wind. As a result, the compounds can travel long distances where they become part of the rain, sleet, snow, and fog as acids.
Equation of reaction:
NO2 + H2O ==> HNO3
NO + H2O ==> HNO2
Final answer:
Nitrogen oxides (NO and NO2) from automobile exhaust contribute to acid rain primarily through the formation of nitric acid and nitrous acid.
Explanation:
High temperatures in the automobile engine cause nitrogen and oxygen gases from the air to combine to form nitrogen oxides (NO and NO2). These nitrogen oxides in automobile exhaust contribute to acid rain by forming nitric acid and nitrous acid. Nitric acid is formed when nitrogen dioxide (NO2), a highly reactive gas and a major component of nitrogen oxides, reacts with atmospheric water. Nitrous acid forms under similar conditions but is less prevalent. The reaction involving nitrogen oxides, particularly with the presence of water vapor, contributes significantly to the phenomenon of acid rain, which has adverse effects on natural water bodies, soil, and vegetation by altering their chemical composition.
Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1133 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.170 M KOH. Calculate the pH of this solution. Ka(HOCl) = 3.5×10-8 pH =
The pH of the given solution is approximately 12.1.
Let's go through the calculations step by step:
Step 1: Moles of Hydrogen Ions in Sulfuric Acid Solution (a)
a = Molarity × Volume
a = 0.100 M × 0.050 L
a = 0.005 mol
Step 2: Moles of Hydrogen Ions in HOCl Solution (b)
Dissociation: HOCl ↦ H+ + OCl-
Ka = [H+][OCl-]/[HOCl] = 3.5 × 10^-8
At equilibrium: (0.1133 - x) M (x) M
x^2/(0.1133 - x) = 3.5 × 10^-8
Solving for x gives x ≈ 1.87 × 10^-4 mol
b = moles of H+ in HOCl solution = x
Step 3: Total Moles of Hydrogen Ions
Total moles of H+ = a + b
Step 4: Moles of Hydroxide Ions in NaOH Solution (c)
c = Molarity × Volume
c = 0.200 M × 0.025 L
c = 0.005 mol
Step 5: Moles of Hydroxide Ions in Ba(OH)2 Solution (d)
d = Molarity × Volume
d = 0.100 M × 0.025 L
d = 0.0025 mol
Step 6: Moles of Hydroxide Ions in KOH Solution (e)
e = Molarity × Volume
e = 0.170 M × 0.010 L
e = 0.0017 mol
Step 7: Total Moles of Hydroxide Ions
Total moles of hydroxide ions = c + d + e
Step 8: Determine Excess Ions (Hydrogen or Hydroxide)
Since Total moles of H+ < Total moles of hydroxide ions, there is excess hydroxide ions.
Step 9: Moles of Excess Hydroxide Ions
Moles of hydroxide left after neutralization = Total moles of hydroxide ions - Total moles of H+
Step 10: Concentration of Hydroxide Ions
Concentration of hydroxide ions left in the solution = Moles of hydroxide left/Total volume of solution
Step 11: Calculate pOH
pOH = -log(OH- concentration)
Step 12: Calculate pH
pH = 14 - pOH
Let's substitute the values and calculate:
pOH = -log(0.001698/0.140) ≈ 1.9
pH = 14 - 1.9 ≈ 12.1
So, the pH of the solution is approximately 12.1.
The calculated pH of the solution is approximately 12.09.
Let's start with the calculations:
1. Calculate Moles of Each Component
H₂SO₄: Moles = 0.050 L * 0.100 M = 0.005 moles of H₂SO₄.HOCl: Moles = 0.030 L * 0.1133 M = 0.003399 moles of HOCl.NaOH: Moles = 0.025 L * 0.200 M = 0.005 moles of NaOH.Ba(OH)₂: Moles = 0.025 L * 0.100 M * 2 = 0.005 moles of OH⁻ (since Ba(OH)₂ dissociates to give 2 OH⁻).KOH: Moles = 0.010 L * 0.170 M = 0.0017 moles of KOH.2. Determine Neutralization
Strong acids (H₂SO₄) and bases (NaOH, KOH, Ba(OH)₂) will neutralize each other:
Total moles of OH⁻ from NaOH, Ba(OH)₂, and KOH = 0.005 + 0.005 + 0.0017 = 0.0117 moles of OH⁻.Total moles of H⁺ from H₂SO₄ = 0.005 * 2 = 0.01 moles (since each H₂SO₄ dissociates to provide 2 H⁺).Neutralization reaction: 0.0117 moles OH⁻ neutralizes 0.01 moles H⁺, leaving 0.0017 moles of OH⁻.3. Calculate Final pH
Since 0.0017 moles of OH- remain:
Total volume of the solution = 50.0 + 30.0 + 25.0 + 25.0 + 10.0 = 140.0 mL = 0.140 L.Concentration of OH⁻ = 0.0017 moles / 0.140 L = 0.01214 M.[H+] = Kw / [OH⁻] = 1.0 × 10-14 / 0.01214 M ≈ 8.24 × 10⁻¹³ M.pH = -log[H⁺] ≈ -log(8.24 × 10⁻¹³) ≈ 12.09.The resulting pH of the mixed solution is approximately 12.09.
53. Consider an electrochemical cell made with zinc in zinc sulfate and copper in copper (II) sulfate. Identify items a through h, for h determine the standard cell potential given that the standard reduction potential for Zn2+ is - 0.763 V and for Cu2+ is + 0.337 V (16 points for answers a – h, and 9 points for i - k):
Answer: Ecell = -0.110volt
Explanation:
Zn--->Zn^+2 + 2e^-.........(1) oxidation
Cu^2+ 2e^- --->Cu........(2)reduction
Zn + Cu^2+ ----> Cu + Zn^+2 (overall
For an electrochemical cell, the reduction potential set up is given by
E(cell) = E(cathode) - E(anode)
E(cell) = E(oxidation) - E(reduction)
E(cathode) = E(oxidation)
E(anode) = E(reduction)
Given that
E(oxidation) = -0.763v
E(reduction) = +0.337v
E(cell) = -0.763 - (+0.337)
E(cell) = -0.763- 0.337
E(cell) = -0.110volt
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+K2CO3(aq)→2KBr(aq)+CO2(g)+H2O(l)
To neutralize the acid in the given equation, you would need 0.1269 grams of K2CO3.
Explanation:To determine the number of grams of K2CO3 needed to neutralize the acid in the given equation, we need to use stoichiometry. First, we determine the molar ratio between K2CO3 and HBr, which is 1:2 based on the balanced equation. Then, we convert the given mass of HBr to moles using its molar mass. Finally, we use the molar ratio and the molar mass of K2CO3 to calculate the mass of K2CO3 needed.
Here's the step-by-step calculation:
Calculate the molar mass of HBr: H: 1.01 g/mol, Br: 79.90 g/mol. Molar mass of HBr = 1.01 g/mol + 79.90 g/mol = 80.91 g/mol.Convert the given mass of HBr to moles: moles of HBr = mass of HBr / molar mass of HBr = 0.1488 g / 80.91 g/mol = 0.001838 mol of HBr.Use the molar ratio between K2CO3 and HBr: 1 mol of K2CO3 : 2 mol of HBr.Calculate the moles of K2CO3 needed: moles of K2CO3 = 0.001838 mol of HBr × (1 mol of K2CO3 / 2 mol of HBr) = 0.000919 mol of K2CO3.Convert the moles of K2CO3 to grams: mass of K2CO3 = moles of K2CO3 × molar mass of K2CO3 = 0.000919 mol of K2CO3 × 138.21 g/mol = 0.1269 g of K2CO3.Therefore, you would need 0.1269 grams of K2CO3 to neutralize the acid according to the given equation.
Learn more about stoichiometry here:https://brainly.com/question/34828728
#SPJ12
To neutralize the acid according to the reaction 2HBr(aq) + K₂CO₃(aq) → 2KBr(aq) + CO₂(g) + H₂O(l), calculate the moles of HBr and use stoichiometry to find the required moles of K₂CO₃. Convert these moles to grams to determine the amount needed.
Chemical Reaction and Calculation:
To determine the grams of K₂CO₃ required to neutralize the acid according to the equation:
2HBr(aq) + K₂CO₃(aq) → 2KBr(aq) + CO₂(g) + H₂O(l)
we need to follow these steps:
Calculate the moles of HBr involved.Use the stoichiometry of the balanced chemical equation to find the moles of K₂CO₃ required.Convert the moles of K₂CO₃ to grams.For a specific example, if you have 0.50 moles of HBr, the balanced equation shows a 2:1 molar ratio between HBr and K₂CO₃. Therefore, you would require 0.25 moles of K₂CO₃. Since the molar mass of K₂CO₃ is 138.205 g/mol, you would need:
0.25 moles × 138.205 g/mol = 34.55 grams of K₂CO₃
10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is initially at 20 c what is the final temperature of the coffee
Answer: The final temperature of the coffee is 43.9°C
Explanation:
To calculate the final temperature, we use the equation:
[tex]q=mC(T_2-T_1)[/tex]
where,
q = heat released = [tex]1.0\times 10^3J=1000J[/tex]
m = mass of water = 10.0 grams
C = specific heat capacity of water = 4.184 J/g°C
[tex]T_2[/tex] = final temperature = ?
[tex]T_1[/tex] = initial temperature = 20°C
Putting values in above equation, we get:
[tex]1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC[/tex]
Hence, the final temperature of the coffee is 43.9°C
Final answer:
The final temperature of the coffee, after adding 1.0 x 10° J of heat to 10.0 g of water initially at 20°C, is approximately 44°C, calculated using the heat transfer formula in thermodynamics.
Explanation:
The subject of this question is Physics, specifically the concept of energy transfer through heating in thermodynamics. To calculate the final temperature of the prepared coffee, we can use the formula q = mcΔT, where q is the heat added, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature. By rearranging the formula to solve for ΔT, we get ΔT = q / (mc).
Given that 1.0 x 103 J of heat is added to 10.0 g of water initially at 20°C, we can calculate ΔT as follows:
ΔT = (1.0 x 103 J) / (10.0 g * 4.18 J/g°C) = 23.923°C (approximately)
Therefore, the final temperature of the water (or coffee) is 20°C + 23.923°C = 43.923°C, which when rounded off, gives us approximately 44°C.
This element and its alloys are used in pumps, valves, and other components that are in contact with acid and petroleum solutions. Group of answer choices
Lead
Zirconium
Nickel
Zinc
Answer:
Nickel
Explanation:
This element and its alloys are used in pumps, valves, and other components that are in contact with acid and petroleum solutions
Final answer:
Nickel and its alloys are used in components like pumps and valves that are exposed to acid and petroleum solutions due to their corrosion resistance.
Explanation:
The element and its alloys mentioned in the question that are used in pumps, valves, and other components in contact with acid and petroleum solutions is likely to be Nickel. This is because nickel and its alloys are known for their corrosion resistance and are used in environments that would rapidly degrade other metals. For instance, nickel is used in the desalination of seawater and nickel steel is used for manufacturing armor plates and burglar-proof vaults. Alloys such as brass (copper and zinc) and bronze (copper, tin, and sometimes zinc) are also important but are not specific to the context given in the question, which highlights usage in acid and petroleum solutions where nickel's properties are most relevant.
Find the equilbrium expression (Ka) for the ionization reaction. HCO3-(aq) + H2O(l) ⇆H3O+(aq) + CO32-(aq) Give the expressions for A, B, and C, given the form:
The equilibrium expression (Ka) for the ionization of bicarbonate ion is given by the formula [H3O+][CO32-] / [HCO3-], where A, B, and C are the equilibrium concentrations of H3O+, CO32-, and HCO3- respectively.
Explanation:To find the equilibrium expression (Ka) for the ionization reaction of the bicarbonate ion (HCO3-(aq)), we must consider the reaction, HCO3-(aq) + H2O(l) ⇌ H3O+(aq) + CO32-(aq). The expression for Ka, known as the acid-ionization constant, follows the general form:
Ka = [H3O+][CO32-] / [HCO3-]
For the reaction given:
A represents the concentration of H3O+B represents the concentration of CO32-C represents the concentration of HCO3-The reaction quotient will be Ka = [A][B] / [C]. Utilizing the values provided, assuming that the concentrations of the products [A] and [B] are equivalent as they form in a 1:1 ratio, and given that we're focusing on the ionization equilibrium of bicarbonate ion as a weak acid, the Ka value can be derived.
Equilibrium expression [tex]\( K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \)[/tex]. Concentrations at equilibrium: [tex]\([A] = [HCO_3^-]\), \([B] = [H_3O^+]\), \([C] = [CO_3^{2-}]\)[/tex].
The equilibrium expression (Ka) for the ionization reaction of [tex]HCO_3^-[/tex](aq) in water is given by the following general formula for an acid dissociation constant:
[tex]\[ K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \][/tex]
Here, the concentration of water ([tex]H_2O[/tex]) is not included in the expression because it is the solvent and its concentration remains relatively constant during the reaction.
For the given reaction:
[tex]\[ HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq) \][/tex]
The expressions for the concentrations of A ([tex]HCO_3^-[/tex]), B ([tex]H_3O^+[/tex]), and C [tex](CO_3^{2-})[/tex] at equilibrium would be:
[tex]\[ [A] = [HCO_3^-] \] \[ [B] = [H_3O^+] \] \[ [C] = [CO_3^{2-}] \][/tex]
These expressions represent the molar concentrations of the species A, B, and C at equilibrium.
The equilibrium constant expression [tex]\( K_a \)[/tex] is then derived from these concentrations, with the product of the concentrations of B and C divided by the concentration of A, as shown in the initial equation.