Answer:
[tex]\large \boxed{0.0106}[/tex]
Explanation:
We have three equations:
1. M(g) ⇌ Z(g); Kc₁ = 3.15
2. 6R(g) ⇌ 2N(g) + 4Z(g); Kc₂ = 0.509
3. 3X(g) + 3Q(g) ⇌ 9R(g); Kc₃ = 12.5
From these, we must devise the target equation:
4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = ?
The target equation has Q(g) on the left, so you divide Equation 1 by 3.
When you divide an equation by 3, you take the cube root of its Kc.
5. X(g) + Q(g) ⇌ 3R(g): K₅ = ∛(Kc₃)
Equation 5 has 3R on the right, and that is not in the target equation.
You need an equation with 3R on the left, so you divide Equation 2 by 2.
When you divide an equation by 2, you take the square root of its Kc.
6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √ (Kc₂)
Equation 6 has 2Z on the right, and that is not in the target equation.
You need an equation with 2Z on the left, so you reverse Equation 2 by and double it.
When you reverse an equation, you take the reciprocal of its K.
When you double an equation, you square its K.
7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²
Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you multiply their K values.
You get the target equation 4:
5. X(g) + Q(g) ⇌ 3R(g); K₅ = ∛(Kc₃)
6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √(Kc₂)
7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²
4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = K₅K₆K₇ = [∛(Kc₃)√(Kc₂)]/(Kc₁)²
Kc = [∛(12.5)√(0.509)]/(12.5)² = (2.321 × 0.7120)/156.2 = 0.0106
[tex]K_{c} \text{ for the reaction is $\large \boxed{\mathbf{0.0106}}$}[/tex]
Answer:
The value of the equilibrium constant is 0.167
Explanation:
Step 1: The target equation
Q(g) + X(g) ⇔ 2M(g) + N(g)
Given is:
(1) M(g)⇔Z(g) c1=3.15
(2) 6R(g) ⇔ 2N(g) + 4Z(g) c2=0.509
(3) 3X(g) +3Q(g) ⇔ 9R(g) c3=12.5
Step 2: Rearange the equation
We have to rearange the equation to come to the final result
This is Hess' Law
In the target equation we have Q(g) + X(g)
In (3) we have 3X(g) +3Q(g) ⇔ 9R(g)
To get the target of Q(g) + X(g) we have to divide (3) by 3. This will give us:
X(g) +Q(g) ⇔ 3R(g) Kc = ∛12.5 = 2.32 (Note: to get Kc of the target equation we use cube root)
The target equation has as product 2M(g) + N(g)
To get M(g) we will use the (1) equation
Since M(g) is a product and not a reactant, we have to reverse the equation. Next to that we also have to double the equation because we need 2M(g) and not M(g)
2Z(g) ⇔ 2M(g) Kc = 1/(3.15)² = 0.101 (Note: to get Kc' after reversing the equation we calculate 1/Kc. To get Kc'' after doubling and reversing the equation we calculate 1/(Kc²)
To get N(g) we will use (2) 6R(g) ⇔ 2N(g) + 4Z(g)
Since we only need N(g) we will divide this equation by 2. This will get us:
3R(g) ⇔ N(g) + 2Z(g) Kc = √0.509 = 0.713 (Note: if we divide the equation by 2, to calculate Kc' we use square root)
Now we have all the components we will add the 3 equations:
X(g) +Q(g) + 2Z(g) + 3R(g)⇔ 3R(g) + 2M(g) + N(g) + 2Z(g)
We will simplify this equation:
X(g) +Q(g) ⇔ 2M(g) + N(g) this is our target equation
The value of the equilibrium constant, Kc is:
Kc = 2.32 * 0.101*0.713
Kc = 0.167
Note: to calculate Kc after adding several equations,we'll multiply Kc1* Kc2 * Kc3 etc...
The value of the equilibrium constant is 0.167
What does the pKa of the conjugate acid of a base have to be to remove a hydrogen from water (pKa = 15.7) so that the ratio of hydroxide (OH-) to water is 100 to 1?
Answer:
The pKa of the conjugate acid is 17.7
Explanation:
If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,
Kₐ = [OH⁻]/[H₂O]
[tex]K_a=\frac{100}{1} =100[/tex]
Now, we determine the equivalent pKa
pKa = -log[ka]
pKa = -log[100]
pKa = -2
Removal of hydrogen from water is reversible as shown below;
H₂O ⇄ OH⁻ + H⁺
15.7 -2
This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];
pKa of the conjugate acid = 15.7 - (-2) = 17.7
The pKa of the conjugate acid is 17.7
Provide a structure for the compound with molecular formula C9H12 and with the following spectroscopic data.1H NMR: 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), 7.1δ (multiplet, I=5H)
Answer:
The structure analysis says the compound must be Cumene or isopropylbenzene
Explanation:
Degree of unsaturation or double bond equivalent
D.B.E = [tex]C-\frac{H}{2}+\frac{N}{2}+1[/tex]
= [tex]9-\frac{12}{2}+\frac{0}{2}+1[/tex]
= 4
¹H NMR data analysis
(i) 1.2δ (doublet, I = 6H) two CH₃ are equivalents and the multiplicity says the neighboring carbon have one hydrogen.
(ii) 3.0δ (septet, I = 1H), one CH and the multiplicity says the neighboring carbon have six hydrogens.
(iii) 7.1δ (multiplet, I = 5H) , means
and the sturcture of the compound is
Final answer:
Based on the 1H NMR spectroscopic data, the structure of the compound with the molecular formula C9H12 is determined to likely be ethylbenzene, featuring an aromatic ring with an attached ethyl group.
Explanation:
The question involves determining the structure of a compound with a molecular formula of C9H12 based on its 1H NMR spectroscopic data. The data provided are 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), and 7.1δ (multiplet, I=5H). To solve this, we analyze each piece of information. The doublet at 1.2δ with 6 hydrogens suggests the presence of two identical methyl groups (CH3) next to a carbon that splits their signal into a doublet. The septet at 3.0δ indicates a methine group (CH) that is adjacent to six hydrogens, likely from two methyl groups, causing this splitting pattern. Finally, the multiplet at 7.1δ for 5 hydrogens indicates the presence of an aromatic ring. Taking all this into account, a plausible structure for this compound is ethylbenzene, which consists of a benzene ring with an ethyl group attached to it.
LeFinal answer:
Based on the 1H NMR spectroscopic data, the structure of the compound with the molecular formula C9H12 is determined to likely be ethylbenzene, featuring an aromatic ring with an attached ethyl group.
Explanation:
The question involves determining the structure of a compound with a molecular formula of C9H12 based on its 1H NMR spectroscopic data. The data provided are 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), and 7.1δ (multiplet, I=5H). To solve this, we analyze each piece of information. The doublet at 1.2δ with 6 hydrogens suggests the presence of two identical methyl groups (CH3) next to a carbon that splits their signal into a doublet. The septet at 3.0δ indicates a methine group (CH) that is adjacent to six hydrogens, likely from two methyl groups, causing this splitting pattern. Finally, the multiplet at 7.1δ for 5 hydrogens indicates the presence of an aromatic ring. Taking all this into account, a plausible structure for this compound is ethylbenzene, which consists of a benzene ring with an ethyl group attached to it.
In the laboratory, a general chemistry student measured the pH of a 0.426 M aqueous solution of hypochlorous acid to be 3.897. Use the information she obtained to determine the Ka for this acid. Ka(experiment)
Answer:
The value of the [tex]K_a[/tex] of the acid is :
[tex]K_a=4.109\times 10^{-8}[/tex]
Explanation:
The pH of the weak acid solution = 3.879
Concentration of hydrogen ions = [tex][H^+][/tex]
[tex]pH=-\log[H^+][/tex]
[tex]3.879=-\log[H^+][/tex]
[tex][H^+]=10^{-3.876}=0.0001321 M[/tex]
Concentration of hypochlorous acid solution = [HClO]= 0.426 M
[tex]HClO(aq)\rightarrow H^+(aq)+ClO^-(aq)[/tex]
Initially:
0.426 M 0 0
At equilibrium
(0.426-0.0001321)M 0.0001321 M 0.0001321 M
The expression of [tex]K_a[/tex] will be given as;
[tex]K_a=\frac{[H^+][ClO^-]}{[HClO]}[/tex]
[tex]K_a=\frac{0.0001321 M\times 0.0001321 M}{(0.426-0.0001321) M}[/tex]
[tex]K_a=4.109\times 10^{-8}[/tex]
The value of the [tex]K_a[/tex] of the acid is :
[tex]K_a=4.109\times 10^{-8}[/tex]
The smell of fresh cut pine is due in part to the cyclic alkene called pinene. Given the following data of pinene: Vapor pressure (torr) Temperature (K) 760 429 515 415 Calculate the heat of vaporization, ΔHvap, of pinene.
The heat of vaporization, ΔHvap, of pinene is 41094 Joules.
Clausius Clapeyron equation:Here the following formula should be used.
ln(P2/P1) = ΔHvap/R (1/T1 - 1/T2)
Here,
P1 = initial pressure at 429 K = 760 torr
P2 = final pressure at 415 K = 515 torr
R = gas constant = 8.314 J/mole.K
T1 = initial temperature = 429 K
T2 = final temperature = 515 K
So, the heat should be
log(515/760) = ΔH/2.303*8.314 {1/429k - 1/415k)
= 41094 J
Hence, The heat of vaporization, ΔHvap, of pinene is 41094 Joules.
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The heat of vaporization (ΔHvap) of pinene is calculated to be approximately 40.3 KJ/mol.
To calculate the heat of vaporization [tex](\(\Delta H_{\text{vap}}\))[/tex] of pinene, we can use the Clausius-Clapeyron equation, which relates the vapor pressure and temperature to the heat of vaporization:
[tex]\[\ln \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\][/tex]
Where:
- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the vapor pressures at temperatures [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex], respectively.
- R is the universal gas constant (8.314 J/mol·K).
- [tex]\( \Delta H_{\text{vap}} \)[/tex] is the heat of vaporization.
Given data:
- [tex]\( P_1 = 760 \)[/tex] torr at [tex]\( T_1 = 429 \)[/tex] K
- [tex]\( P_2 = 515 \)[/tex] torr at [tex]\( T_2 = 415 \)[/tex] K
First, we convert the pressures to the same units if necessary, but here they are both in torr.
Now, let's use the Clausius-Clapeyron equation to solve for [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[\ln \left( \frac{760}{515} \right) = \frac{\Delta H_{\text{vap}}}{8.314} \left( \frac{1}{415} - \frac{1}{429} \right)\][/tex]
Calculate the natural logarithm and the reciprocal temperatures:
[tex]\[\ln \left( \frac{760}{515} \right) = \ln(1.4757) \approx 0.388\][/tex]
[tex]\[\frac{1}{415} \approx 0.00241 \quad \text{K}^{-1}\][/tex]
[tex]\[\frac{1}{429} \approx 0.00233 \quad \text{K}^{-1}\][/tex]
[tex]\[\frac{1}{415} - \frac{1}{429} \approx 0.00241 - 0.00233 = 0.00008 \quad \text{K}^{-1}\][/tex]
Now plug these values back into the Clausius-Clapeyron equation:
[tex]\[0.388 = \frac{\Delta H_{\text{vap}}}{8.314} \times 0.00008\][/tex]
Solve for [tex]\(\Delta H_{\text{vap}}\):[/tex]
[tex]\[\Delta H_{\text{vap}} = \frac{0.388 \times 8.314}{0.00008}\][/tex]
[tex]\[\Delta H_{\text{vap}} = \frac{3.226}{0.00008}\][/tex]
[tex]\[\Delta H_{\text{vap}} \approx 40325 \text{ J/mol} = 40.3 \text{ kJ/mol}\][/tex]
So, the heat of vaporization [tex](\(\Delta H_{\text{vap}}\))[/tex] of pinene is approximately 40.3 kJ/mol.
The temperature on a distant, undiscovered planet is expressed in degrees B. For example, water boils at 180 ∘ B and freezes at 50 ∘ B . If it is 31 ∘ C on Earth, what would that temperature be in terms of degrees B?
To convert a temperature of 31 degrees Celsius to the equivalent temperature in degrees B on the undiscovered planet, we must use the given ratios of temperature differences, resulting in a converted temperature of 90.3 degrees B.
Explanation:The temperature range for water between freezing and boiling on this undiscovered planet expressed in degrees B is 130 (180-50). The same range on Earth in degrees Celsius is 100 (100-0). First, we need to find the ratio of these two scales. The ratio is 130/100 = 1.3.
Next, to convert the Earth temperature from Celsius (31 °C) to degrees B, we multiply by the ratio and add the freezing point in degrees B. This gives us: (31 * 1.3) + 50 = 90.3 °B. So, 31 °C on Earth would be 90.3 °B on this undiscovered planet.
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Ethane, C2H6, burns in oxygen. First write a balanced equation for this combustion reaction. What mass of oxygen, in grams, is required for complete combustion of 13.6 g of ethane
Answer:
We need 50.6 grams of oxygen
Explanation:
Step 1: Data given
Mass of ethane = 13.6 grams
Molar mass of ethane = 30.07 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles ethane
Moles ethane = mass ethane / molar mass ethane
Moles ethane = 13.6 grams / 30.07 g/mol
Moles ethane = 0.452 moles
Step 4: Calculate moles oxygen
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
For 0.452 moles ethane we need 3.5*0.452 = 1.582 moles O2
Step 5: Calculate mass O2
Mass O2 = moles O2 * molar mass O2
Mass O2 = 1.582 moles * 32.0 g/mol
Mass O2 = 50.6 grams
We need 50.6 grams of oxygen
The balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O. To burn 13.6g of ethane completely, approximately 50.63g of oxygen is needed.
Explanation:Firstly, the balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
From this equation, one mole of ethane reacts with 3.5 moles of oxygen. To calculate the mass of oxygen needed, you firstly need to know the molar mass of ethane (C2H6), which is approximately 30.07 g/mol. Therefore, 13.6 g of ethane is about 0.452 moles.
Since 1 mole of ethane reacts with 3.5 moles of oxygen, then 0.452 moles of ethane would require (0.452 x 3.5) = 1.582 moles of oxygen. The molar mass of oxygen (O2) is about 32 g/mol, so the mass of oxygen required is (1.582 moles x 32 g/mol) = 50.63 g.
Therefore, the mass of oxygen required for the complete combustion of 13.6 g of ethane is approximately 50.63 grams.
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All of the following processes lead to an increase in entropy EXCEPT A) increasing the temperature of a gas. B) melting a solid. C) chemical reactions that increase the number of moles of gas. D) forming mixtures from pure substances. E) decreasing the volume of a gas.
Answer: D) forming mixtures from pure substances.
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) Increase in temperature of a gas: As increasing the temperature , increases the kinetic energy of molecules , the molecules move faster and thus entropy increases.
b) Melting a solid : The randomness will increase as liquids move freely as compared to solids and hence entropy will also increase.
c) chemical reactions that increase the number of moles of gas: As more gaseous molecules will be formed, more will be the randomness and hence entropy increases.
d) forming mixtures from pure substances: As substances in a mixture do not react chemically and thus the molecules remain same and entropy remain same.
e) decreasing the volume of a gas: According to Boyle's law, decreasing the volume will increase the pressure and thus entropy will increase.
Final answer:
The process that does not lead to an increase in entropy is decreasing the volume of a gas as it compact the molecules into a smaller space, reducing disorder.
Explanation:
The question revolves around identifying which process does not lead to an increase in entropy. Entropy is a measure of the disorder or randomness in a system. Several processes can increase entropy, such as:
Increasing the temperature of a gas, which increases the kinetic energy and disorder of the gas molecules.
The process of melting a solid, where the ordered solid structure becomes a more disorderly liquid.
Chemical reactions that increase the number of gas molecules, as a greater number of particles usually means greater disorder.
Forming mixtures from pure substances, whereby the uniform structure becomes more randomly arranged upon mixing.
However, the one process that does not increase entropy is decreasing the volume of gas. Decreasing the volume of a gas reduces its entropy because it compacts the gas molecules into a smaller space, potentially giving the system less disorder.
Therefore, the correct answer is E) decreasing the volume of a gas.
Morphine has the formula C17H19NO3. It is a base and accepts one proton per molecule. It is isolated from opium. A 0.682g sample of opium is found to require 8.92 mL of a 0.0116 M solution of sulfuric acid for neutralization. Assuming that morphine is the only base present in opium, calculate the mass (in grams) of morphine in the sample of opium.
Reaction equation: 2 C17H19NO3 + H2SO4 --> Product
Answer: The mass of morphine is 0.059 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of sulfuric acid solution = 0.0116 M
Volume of solution = 8.92 mL
Putting values in above equation, we get:
[tex]0.0166M=\frac{\text{Moles of sulfuric acid}\times 1000}{8.92}\\\\\text{Moles of sulfuric acid}=\frac{0.0116\times 8.92}{1000}=1.035\times 10^{-4}mol[/tex]
The given chemical equation follows:
[tex]2C_{17}H_{19}NO_3+H_2SO_4\rightarrow \text{Product}[/tex]
By Stoichiometry of the reaction:
1 moles of sulfuric acid reacts with 2 moles of morphine
So, [tex]1.035\times 10^{-4}mol[/tex] of sulfuric acid will react with = [tex]\frac{2}{1}\times 1.035\times 10^{-4}=2.07\times 10^{-4}mol[/tex]
To calculate the mass of morphine for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of morphine = 285.34 g/mol
Moles of morphine = [tex]2.07\times 10^{-4}mol[/tex]
Putting values in above equation, we get:
[tex]2.07\times 10^{-4}mol=\frac{\text{Mass of morphine}}{285.34}\\\\\text{Mass of morphine}=(2.07\times 10^{-4}mol\times 285.34g/mol)=0.059g[/tex]
Hence, the mass of morphine is 0.059 grams
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degrees celsius . During the reaction 69.0kJ of heat flows out of the bath and into the flask.
Required:
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J*g*K. Round your answer to 3 significant digits.
Answer:
[tex]T_F=31.9^0C[/tex]
Explanation:
Hello,
In this case, the energy content is modeled via:
[tex]Q=mCp(T_F-T_0)[/tex]
Since the unknown is the final temperature, it is solved by:
[tex]T_F=T_0+\frac{Q}{mCp} =33.9^0C+\frac{-69.0kJ}{8.10kg*4.18\frac{kJ}{kg*^0C}} \\T_F=31.9^0C[/tex]
The heat is negative since it flows out of the bath that is the considered system.
Best regards.
The normal freezing point of a certain liquid Y is -8.70°C, but when 24.17g of urea (NH2)2CO are dissolved in 550 g of Y, it is found that the solution freezes at -13.7°C instead. Use this information to calculate the molal freezing point depression constant Kf of substance Y.
Answer:
6.8 ºC m⁻¹
Explanation:
The strategy here is to use the equation for the freezing point depression
ΔTf = Kf x m ⇒ Kf = ΔTf / m
where k is the freezing point depression constant,ΔTf is the change in the freezing point of the solution ( freezing point pure solvent minus freezing point of the solution), and m is the molality, mol of solute per kilogram solvent.
Thus
ΔTf (ºC) = -8.7 ºC - ( -13.7 )C = 5.0 ºC
mol (NH₂)₂CO = mass / molar mass = 24.17 g / 60.06 gmol⁻¹ = 0.40 mol
m = 0.40 mol / 0.550 Kg = 0.73 m
Kf = ΔTf / m = 5.0 ºC / 0.73 m = 6.8 ºCm⁻¹
The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is added to the first beaker and 100. mL of 8 M HCl is added to the second. (a) Find the ion-product constant for KCl at 20 °C. 14 Enter as a number to 2 decimal places. (b) What mass, if any, of KCl will precipitate from each beaker? Enter as a number to 0 decimal places. beaker 1: 0 grams beaker2: grams
Answer:
a)The Ksp was found to be equal to 13.69
Explanation:
Terminology
Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.
Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.
Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).
Step-by-step solution:
To solve this:
#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.
#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.
#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.
a) The equation of solubility equilibrium for KCL is thus;
KCL_(s) ---> K+(aq) + Cl- (aq)
The solubility of KCl given is 3.7 M.
Ksp= [K+][Cl-] = (3.7)(3.7) =13.69
The Ksp was found to be equal to 14.
In pure water KCl
Ksp =13.69 KCl =[K+][Cl-]
Let x= molar solubility [K+],/[Cl-] :. × , x
Ksp =13.69 = [K+][Cl-] = (x)(x) = x²
x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio
37M moles/L
The Ksp was found to be equal to 14.
4.0 M HCl = KCl =[K+][Cl-]
Let y= molar solubility :. y, y+4
Ksp =13.69= [K+][Cl-] = (y)(y*+4)
* - rule of thumb
Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)
13.69=4y:. y= 3.42 moles/100mL
y= 34.2moles/L
8 M HCl = KCl =[K+][Cl-]
Let b= molar solubility :. B, b+8
Ksp =13.69= [K+][Cl-] = (b)(b*+8)
* - rule of thumb
Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)
13.69=8b:. b= 1.71 moles/100mL
17.1 moles/L
Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.
Answer:
(a) 13.69
(b) i beaker 1: 0g
ii beaker 2: 0g
Explanation:
a. The solubility equilibrium equation for KCl is
KCl(s) ⇄ K⁺(aq) + Cl⁻(aq)
3.7M KCl contains equal moles of K ions and Cl ions
therefore, the ion-product expression is written thus
Ksp = [K⁺][Cl⁻]
= [3.7][3.7]
= 13.69
b. from the first two beakers containing 100 mL and 3.7M KCl
moles of K⁺ = moles of Cl⁻ = moles of KCl = 3.7moles in 1L
if 3.7M Implies 3.7 moles in 1L or 1000 mL or 1000 cm³
how many moles will be contained in 100 mL
this is calculated as follows
3.7moles/Liter * 100 mL
[tex]\frac{3.7 moles KCl}{1000 mL} * 100 ml = 0.37moles KCl[/tex]
= 0.37moles K⁺ = 0.37moles Cl⁻
4.0 M HCl, contains
[tex]\frac{4 moles HCl}{1000 mL} *100mL = 0.4moles HCl = 0.4 moles H = 0.4moles Cl[/tex] in 100mL
8.0M HCl, contains
[tex]\frac{8moles HCL}{1000mL} *100mL=0.8mole HCl=0.8molesH=0.8molesCl[/tex] in 100mL
now, in the first beaker 100 mL of 4M HCl is added to 100 mL of 3.7M KCl
total moles of Cl⁻ (0.4 + 0.37) moles = 0.77 moles
total moles of K⁺ remains 0.37 moles
total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L
total moles of Cl⁻ per Liter = 0.77moles/0.2L = 3.85M Cl⁻
total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺
Qsp must be greater or equal to Ksp for Precipitation to occur, that is
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][3.85] = 7.12 this is less than 13.69(Ksp)
hence no KCl will precipitate in the first beaker
since there is no precipitate, there is therefore no need for calculating the mass precipitated
and the answer is 0g
(bii) now, in the second beaker 100 mL of 8M HCl is added to 100 mL of 3.7M KCl
total moles of Cl⁻ (0.8 + 0.37) moles = 1.17 moles
total moles of K⁺ remains 0.37 moles
total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L
total moles of Cl⁻ per Liter = 1.17moles/0.2L = 5.85M Cl⁻
total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺
Qsp must be greater or equal to Ksp for Precipitation to occur, that is
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][5.85] = 10.82 this is less than 13.69(Ksp)
hence no KCl will precipitate also in the second beaker
since there is no precipitate, there is therefore no need fo calculating the mass precipitated
and the answer is 0g
Consider the three isomeric alkanes n-hexane, 2, 3-dimethylbutane, anu 2-llylp following correctly lists these compounds in order of increasing boiling point? Circle correct Answer.
A) 2, 3-dimethylbutane < 2-methylpentane < n-hexane
B) 2-methylpentane
C) 2-methylpentane < 2, 3-dimethylbutane < n-hexane
D) n-hexane<2-methylpentane <2, 3-dimethylbutane
E) n-hexane <2, 3-dimethylbutane<2-methylpentane
Answer:
D) n-hexane<2-methylpentane <2, 3-dimethylbutane
Explanation:
structures with less branching have high boiling point and low melting points this is because of freezing depression and boiling depression and stronger bonds.
To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment involving tiny, charged drops of oil. The drops were charged because they had picked up extra electrons. Millikan was able to measure the charge on each drop in coulombs. Here is an example of what his data may have looked like. Based on the given data, how many extra electrons did drop C contain? Express your answer as an integer.
The question is incomplete, here is the complete question:
To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment involving tiny, charged drops of oil. The drops were charged because they had picked up extra electrons. Millikan was able to measure the charge on each drop in coulombs. Here is an example of what his data may have looked like.
Drop Charge (C)
A -3.20 × 10⁻¹⁹
B -4.80 × 10⁻¹⁹
C -8.00 × 10⁻¹⁹
D -9.60 × 10⁻¹⁹
Based on the given data, how many extra electrons did drop C contain? Express your answer as an integer.
Answer: The extra electrons that the drop C contain are 5
Explanation:
Millikan’s oil drop experiment is used to measure the charge of an electron. Before this experiment, the subatomic particles were not accepted.
He found that all the oil drops had charges that were the multiples of [tex]-1.6\times 10^{-19}C[/tex]. This value is the charge on 1 electron
Number of electrons excess electrons is calculated by using the formula:
[tex]\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}[/tex]
For Drop C:
Charge on drop C = [tex]-8.00\times 10^{-19}C[/tex]
[tex]\text{Excess electrons}=\frac{-8.00\times 10^{-19}}{-1.6\times 10^{-19}}=5[/tex]
Hence, the extra electrons that the drop C contain are 5
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH. Assume no volume change upon the addition of base. The Ka for HF is 6.8 × 10-4.
Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of [tex]HF[/tex].
[tex]\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}[/tex]
[tex]\text{Moles of HF}=0.250M\times 1.50L=0.375mol[/tex]
Now we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (6.8\times 10^{-4})[/tex]
[tex]pK_a=4-\log (6.8)[/tex]
[tex]pK_a=3.17[/tex]
The reaction will be:
[tex]HF+OH^-\rightleftharpoons F^-+H_2O[/tex]
Initial moles 0.375 0.100 0.375
At eqm. (0.375-0.100) 0 (0.375+0.100)
= 0.275 = 0.475
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[F^-]}{[HF]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}][/tex]
[tex]pH=3.41[/tex]
Thus, the pH of the solution is, 3.41
In the presence of a base, the weak acid (HF) in the buffer will react to form its conjugate base (F-), shifting the buffer balance and making the solution increasingly basic. Using the given Ka and the calculated moles, we're able to figure out the new pH after the addition of NaOH.
Explanation:This question revolves around the concept of buffers, particularly acid-base titration. A buffer is a solution that resists changes in its pH when small amounts of a strong acid or a strong base are added. Therefore, when NaOH is added, it will first react with HF, a weak acid. Since HF is a weak acid and F- a weak base, the reaction of NaOH with HF will increase the amount of F-, shifting the pH.
Now, here's how we calculate it. Use the relation Ka = [H+][F-]/[HF], then calculate [H+]. Follow this process:
Calculation of moles before the addition of the NaOH : HF = NaF = 1.5 L * 0.25 M = 0.375 mol Reaction with NaOH: 0.375 mol - 0.1 mol = 0.275 mol. Assuming the NaOH reacts entirely with the HF, the new amounts are: HF = 0.275 mol and NaF = 0.475 mol Next, we know [F-] = [NaF]/Volume = 0.475 mol /1.5 L = 0.317 M Then, Ka = [H+][F-]/[HF] gives us [H+] = Ka * [HF] / [F-] = 6.8×10⁻⁴ * ([0.275 mol]/1.5 L) / ([0.475 mol]/1.5L) Finally, calculate the pH by using pH = -log[H+]Learn more about Buffer Solutions here:https://brainly.com/question/31367305
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The net ionic equation, H3PO4 (aq) + 3 OH− (aq) Imported Asset PO4−3 (aq) + 3 H2O (l), best represents which type of acid-base reaction?
strong acid-strong base
weak acid-strong base
strong acid-weak base
weak acid-weak base
Answer:
Weak acid - strong base
Explanation:
H₃PO₄ → Phosphoric acid.
This is a weak that has three dissociations in order to give hydronium to the medium and to produce the phosphate anion. The equations are:
H₃PO₄ + H₂O ⇄ H₃O⁺ + H₂PO₄⁻ Ka1
H₂PO₄⁻ + H₂O ⇄ H₃O⁺ + HPO₄⁻² Ka2
HPO₄⁻² + H₂O ⇄ H₃O⁺ + PO₄⁻₃ Ka3
As the H₃PO₄ is a weak acid then the water behaves as a strong base.
If we follow the Brownsted Lory idea, water becomes a strong base cause it receives the H⁺ from water, then the H₃O⁺ becomes the conjugate weak acid.
Anions from the H₃PO₄, diacid phosphate and monoacid phosphate assume the rol of the conjugate strong base, they all recieve proton but this is a special case, because both anions can recieve all release the proton. That's why, they also are amphoteric
A 62.5-g piece of gold at 650. K is dropped into 165 g of H2O (l) at 298 K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP,m for Au and H2O are constant at their values for 298 K throughout the temperature range of interest.
Final answer:
To find the final temperature of the system when a gold piece is dropped into water, apply the conservation of energy principle and equate the heat lost by gold to the heat gained by water using their specific heat capacities.
Explanation:
A 62.5-g piece of gold at 650 K is dropped into 165 g of H₂O (l) at 298 K in an insulated container at 1 bar pressure.
To find the final temperature of the system, you can use the principle of conservation of energy, which states that the heat lost by the gold will be gained by the water.
Calculate the energy transferred using the specific heat capacities of gold and water along with the heat gained or lost formula, then set the heat lost by gold equal to the heat gained by water to find the final equilibrium temperature.
g In Part 7, the [Cl-] in saturated NaCl is 5.4 M at room temperature. Assume that you had 1.00 ml of the saturatedsolution, and that you added 0.50 ml of 12 M HCl. What is the [Cl-] after you added the HCl. (When two solutionscontain the same component, the numerator consists of the sum of the volume times the concentration for each solu-tion. The denominator is the total volume.
Answer:
7.60 M
Explanation:
Our method to solve this question is to use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have
[Cl⁻] = # mol Cl⁻ / Vol
Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:
1mL x 1 L / 1000 mL = 0.001 L
0.5 mL x 1L/1000 mL = 0.0005 L
[Cl⁻] = 0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )
= 7.6 M
This is the same as the statement given in the question.
When solid (NH4)(NH2CO2)(NH4)(NH2CO2) is introduced into and evacuated flask at 25∘C25∘C, the total pressure of gas at equilibrium is 0.30.3 atm. What is the value of Kpat25∘CKpat25∘C?
Answer:
Kp = 0.004 = 4*10^-3
Explanation:
Step 1: Data given
Temperature = 25.0 °C
the total pressure of gas at equilibrium is 0.30 atm
Step 2: The balanced equation
NH4(NH2CO2)(s) ⇆ 2NH3(g) + CO2(g)
For 1 mol NH4(NH2CO2) we have 2 moles NH3 and 1 mol CO2
Step 3: The total pressure
Total pressure = pNH3 + pCO2
Total pressure = 2X + X
Total pressure = 3X = 0.30
X = 0.10
Step 4: The partial pressures
pNH3 = 2X = 2*0.10 = 0.20 atm
pCO2 = X = 0.10 atm
Step 5: Calculate Kp
Kp = (pNH3)²*(pCO2)
Kp = (0.20²)*0.10
Kp = 0.004 = 4*10^-3
The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equivalence point and 9.27 at the second equivalence point. If the add solution contained 0.2015 g of the acid, what is the molar mass, pK_a1, and pK_a2 of the acid?
Answer:
molar mass (of the diprotic acid) = 116g/mol
[tex]pK_{a_1}= 1.83[/tex]
[tex]pK_{a_2}=6.07[/tex]
Explanation:
The number of moles of [tex]NaOH[/tex] and [tex]H^+[/tex] are equal at the equivalence point since they are both taking part in the diprotic acid.
0.1 M means 0.1 moles in 1L or 0.1 moles in 1000 mL
Number of moles of 0.1 M NaOH at final equivalence point ;
= [tex]34.72 mL * \frac{0.1mole}{1000mL}[/tex]
= 0.00347 moles of NaOH
However, the number of moles of the diprotic acid in the 0.25 L solution is = [tex]\frac {0.00347}{2 }[/tex] (due to the fact that half of the concentration of NaOH is needed to give the same amount of [tex]H^+[/tex]
[tex]= 0.001736[/tex]
Given that:
The Total acid in the solution = 0.2015 g; to calculate the molar mass ; we have :
no of moles = [tex]\frac{mass}{molar mass}[/tex]
molar mass= [tex]\frac{mass}{numbers of moles}[/tex]
molar mass = [tex]\frac{0.2015}{0.001736}[/tex]
molar mass (of the diprotic acid) = 116g/mol
At the second equivalence point;
The pH = 9.27
pH = [tex]-log[H^+][/tex]
[tex][H^+]= 10^{-pH}[/tex]
[tex][H^+] = 10^{-9.27[/tex]
[tex][H+] = 5.37 * 10^{-10[/tex]
[tex][OH^-][/tex] can be calculated as follows:
[tex]\frac{10^{-14}}{[H^+]} = \frac {10^{-14}}{5.37* 10^{-10}}[/tex]
[tex][OH^-][/tex] = [tex]1.86*10^{-5}[/tex]
Let represent the equation from the reaction after the second equivalence point with:
[tex]X^{2-} + H_2O \rightleftharpoons HX^- + OH^-[/tex]
where:
[tex]X^{2-}[/tex] = [tex]\frac{0.001736}{(25 + 34.72)} * 1000[/tex]
[tex]X^{2-}[/tex] = [tex]0.029 M[/tex]
The ICE Table is shown as follows;
[tex]X^{2-}[/tex] [tex]+[/tex] [tex]H_2O[/tex] [tex]\rightleftharpoons[/tex] [tex]HX^-[/tex] [tex]+[/tex] [tex]OH^-[/tex]
Initial 0.029 0 0
Change -x +x +x
Equilibrium (0.029 - x) x x
[tex]Kb = \frac{[HX^-][OH^-]}{[X^{2-}]}[/tex]
[tex]Kb = \frac{x^2}{0.029-x}[/tex]
[tex]k_b}= \frac{(1.86*10^{-5})^2}{0.029-(1.86*10^{-5})}[/tex]
[tex]x = 1.191*10^{-8}[/tex]
[tex]K_a} = \frac {10^{-14}}{K_b}[/tex]
[tex]K_a} = \frac {10^{-14}}{1.191*10^{-8}}[/tex]
[tex]K_a} = 8.39 * 10^{-7}[/tex]
[tex]pK_a} = -log K_a[/tex]
[tex]pK_a} = -log (8.39*10^{-7})[/tex]
[tex]pK_{a_2}=6.07[/tex]
At the first equivalence point we have all H2X getting converted to [tex]HX^-[/tex] [tex]HX^-[/tex] is an amphoteric species which implies that it can serve as both an acid and a base
As such, in this process:
[tex]pH = pK_{a_1} + \frac{pKa_2}{2}[/tex]
Given that: the pH = 3.95
Then;
[tex]3.95 = pK_{a_1} + \frac{6.07}{2}[/tex]
[tex]3.95*2 = pK_{a_1} +{6.07}[/tex]
[tex]7.9 = pK_{a_1} +{6.07}[/tex]
[tex]pK_{a_1}= 7.9 - {6.07}[/tex]
[tex]pK_{a_1}= 1.83[/tex]
Answer:
gfhgfhhgjffhfhgfhgfhfgfhfg
Explanation:
"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volumetric flask. You find that the solution has an osmotic pressure of 531 mm Hg at 25 °C. What is the composition of the mixture?"
Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature = [tex]25^{o} C[/tex]
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 = [tex]C \times 0.082 \times 298.15 K [/tex]
C = 0.0285
This also means that,
[tex]\frac{\text{moles}}{\text{volume (in L)}}[/tex] = 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
= [tex]2.85 \times 10^{-3 }[/tex]
Now, let us assume that mass of [tex]C_{12}H_{23}O_{5}N[/tex] = x grams
And, mass of [tex]C_{12}H{22}O_{11}[/tex] = (1.00 - x)
So, moles of [tex]C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{x}{369}[/tex]
Now, moles of [tex]C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}[/tex]
= [tex]\frac{x}{369} + \frac{(1.00 - x)}{342}[/tex]
= [tex]2.85 \times 10^{-3}[/tex]
= x = 0.346
Therefore, we can conclude that amount of [tex]C_{12}H_{23}O_{5}N[/tex] present is 0.346 g and amount of [tex]C_{12}H_{22}O_{11}[/tex] present is (1 - 0.346) g = 0.654 g.
How many carbon atoms are there in a molecule of each of the following:
a. ethane b. 1-butyne c. propene d. cyclooctane
Please provide molecular formula for each of the four above mentioned compounds.
Answer:
a. 2 carbons; b. 4 carbons; c. 3 carbons; d. 8 carbons
Explanation:
Think in the prefixes
a. Ethane → Eth (2C)
C₂H₆
b. 1-butyne → But (4C)
C₄H₆
c. Propene → Prop (3C)
C₃H₆
d. Cyclooctane → Oct (8C)
C₈H₁₆
Final answer:
In summary, ethane has 2 carbon atoms with the molecular formula C₂H₆, 1-butyne has 4 carbon atoms with the formula C₄H₆, propene has 3 carbon atoms with the formula C₃H₆, and cyclooctane has 8 carbon atoms with the formula C₈H₁₆.
Explanation:
The subject of the question is specifically about determining the number of carbon atoms in various hydrocarbon molecules and providing their molecular formulas:
Ethane: The molecular formula for ethane is C₂H₆, which indicates that there are 2 carbon atoms in an ethane molecule.1-Butyne: The molecular formula for 1-butyne is C₄H₆. It has 4 carbon atoms.Propene: The molecular formula for propene is C₃H₆. It consists of 3 carbon atoms.Cyclooctane: The molecular formula for cyclooctane is C₈H₁₆, indicating it has 8 carbon atoms in the molecule.When H 2SO 4 is neutralized by NaOH in aqueous solution, the net ionic equation is ________. SO42- (aq) 2Na (aq) Na2SO4 (aq) SO42- (aq) 2Na (aq) Na2SO4 (s) H (aq) OH- (aq) H2O (l) H2SO4 (aq) 2OH- (aq) 2H2O (l) SO42- (aq) 2H (aq) 2NaOH (aq) 2H2O (l) 2Na (aq)
Answer:
H+(aq) + OH-(aq) ——> H2O(l)
Explanation:
The net ionic equation for the neutralization reaction between H₂SO₄ and NaOH is
2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l)
From the question,
We are to write the net ionic equation for the neutralization reaction between H₂SO₄ and NaOH
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
Now, we will write the ionic equation
The ionic equation for the reaction is
2H⁺(aq) + SO₄²⁻(aq) + 2Na⁺(aq) 2OH⁻(aq) → 2Na⁺(aq) + SO₄²⁻(aq) + 2H₂O(l)
Then, cancel out the spectator ions, Na⁺ and SO₄²⁻ , to give the net ionic equation
The net ionic equation for the neutralization reaction is
2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l)
Hence, the net ionic equation for the neutralization reaction between H₂SO₄ and NaOH is
2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l)
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A 3.90 g sample of Cl2 reacts with PCl3 to form 11.45 g of PCl5 according to the reaction below. How much PCl3 is needed? [Hint: You do not need to calculate moles to answer.]
PCl3 + Cl2 —> PCl5 Amount of PCl3:
Show work here
Answer:
We need 7.55 grams of PCl3
Explanation:
Step 1: Data given
Mass of Cl2 = 3.90 grams
Mass of PCl5 = 11.45 grams
Molar mass Cl2 = 70.9 g/mol
Molar mass PCl5 = 208.24 g/mol
Step 2: The balanced equation
PCl3 + Cl2 → PCl5
Step 3: calculate moles
Moles = mass / molar mass
Moles Cl2 = 3.90 grams / 70.9 g/mol
Moles Cl2 = 0.0550 moles
Moles PCl5 = 11.45 grams /208.24 g/mol
Moles PCl5 = 0.0550 moles
Step 4: Calculate moles PCl3
For 1 mol PCl3 we need 1 mol Cl2 to produce 1 mol PCl5
For 0.0550 moles PCl5 we will need 0.0550 moles Cl2 and 0.0550 moles PCl3
Step 5: Calculate mass PCl3
Mass PCl3 = moles * molar mass
Mass PCl3 = 0.0550 moles * 137.33 g/mol
Mass PCl3 = 7.55 grams
Without calculating the number of moles:
Mass of Cl2 + mass PCl3 = mass PCl5
Mass PCl3 = 11.45 - 3.90 = 7.55 grams
We need 7.55 grams of PCl3
Final answer:
To find the mass of PCl3 needed for the reaction, use the mass difference between the product PCl5 and the reactant Cl2, resulting in 7.55 g of PCl3.
Explanation:
The question involves a stoichiometry calculation to determine the amount of phosphorus trichloride (PCl₃) required to react with chlorine gas (Cl₂) to produce phosporus pentachloride (PCl₅). If 3.90 g of Cl₂ reacts to form 11.45 g of PCl₅, we can use the principle of conservation of mass to find the mass of PCl₃ that reacted. According to the reaction PCl₃ + Cl₂
PCl₅, for every mole of PCl₅ produced, one mole of PCl₃ is consumed.
In this problem, since we start with Cl₂ and end up with PCl₅, the difference in mass between PCl₅ and Cl₂ must be the mass of PCl₃ consumed. Therefore, the mass of PCl₃ is given by subtracting the mass of Cl₂ from the mass of PCl₅:
Mass of PCl₃ = Mass of PCl₅ - Mass of Cl₂
= 11.45 g - 3.90 g
= 7.55 g
Thus, 7.55 g of PCl₃ is needed for the reaction.
A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode composed of copper in a 1.0 M copper(I) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C .
Answer: The standard potential of the cell is 0.77 V
Explanation:
We know that:
[tex]E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V[/tex]
The substance having highest positive [tex]E^o[/tex] reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq)+2e^-[/tex]
Reduction half reaction: [tex]Cu^{+}(aq)+e^-\rightarrow Cu(s)[/tex] ( × 2)
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:
[tex]E^o_{cell}=0.52-(-0.25)=0.77V[/tex]
Hence, the standard potential of the cell is 0.77 V
Consider the following three processes: (1) Melting of ice at room temperature (2) Boiling of water at 101°C (3) Dissolving of NH4NO3 with water in an instant cold pack? Which statement(s) that are true about ALL three of the given processes? (Choose one or more)(A) Endothermic(B) Exothermic(C) Nonspontaneous(D) Spontaneous(E) none of these statements can be used to describe all three processes
Answer: Option (A) is the correct answer.
Explanation:
Exothermic reaction is defined as the reaction in which release of heat takes place. Whereas endothermic reaction is defined as the reaction in which heat is absorbed by the reactant molecules.
When ice melts at room temperature it is an endothermic reaction as it occurs due to absorption of heat. Boiling of water at [tex]101^{o}C[/tex] is also an endothermic process. This is because heat is absorbed by water molecules due to which their state changes from liquid to vapor form. When [tex]NH_{4}NO_{3}[/tex] is dissolved in water then heat is absorbed as there occurs a decrease in temperature of water. Hence, it is also an endothermic reaction.Thus, we can conclude that for the given statements its is true that ALL three of the given processes are endothermic.
For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of carbon dioxide. benzene (C6H6) (l) oxygen (g) carbon dioxide (g) water (g) What is the theoretical yield of carbon dioxide
Answer:
Theoretical yield for CO₂ is 5.10g
Explanation:
Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)
We convert the mass of oxygen to moles:
4.64 g /32 g/mol = 0.145 moles of O₂
Let's find out the 100% yield reaction of CO₂ (theoretical yield)
Ratio is 15:12. So let's make this rule of three:
15 moles of O₂ can produce 12 moles of CO₂
Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles
We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g
Answer:
The theoretical yield of carbon dioxdide = 5.11 grams
Explanation:
Step 1: Data given
Mass of oxygen gas = 4.64 grams
Molar mass of O2 = 32.0 g/mol
The reaction yields 3.95 grams of carbon dioxide (CO2)
Molar mass CO2 = 44.01 g/mol
Step 2: The balanced equation
2C6H6 + 15O2 → 12CO2 + 6H2O
Step 3: Calculate moles oxygen
Moles oxygen = mass oxygen / molar mass oxygen
Moles oxygen = 4.64 grams / 32.0 g/mol
Moles oxygen = 0.145 moles
Step 4: Calculate moles of carbon dioxide (CO2)
For 2 moles C6H6 we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
For 0.145 moles O2 we'll have 12/15 * 0.145 = 0.116 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.116 moles * 44.01 g/mol
Mass CO2 = 5.11 grams
The theoretical yield of carbon dioxdide = 5.11 grams
A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq ) was back titrated with 23.60 mL of 0.0980 M NaOH ( aq ) . How many moles of HCl react with the carbonate?
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = [tex]\frac{0.1200}{1000}\times 50.0[/tex] moles = 0.00600 moles
Neutralization reaction (back titration): [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = [tex]\frac{0.0980}{1000}\times 23.60[/tex] moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
Calculate the mass % of magnesium sulfate (assume that there is a 1:1 mol ratio between sulfate and magnesium sulfate) in the original sample. Report your answer without units and use 3 sig figs, i.e. 55.23543% would be entered as 55.2
61.8 % is the mass percentage of magnesium sulphate.
Explanation:
The mass percent of individual solute or ion in a compound is calculated by the formula:
Grams of solute ÷ grams of solute + solvent × 100
mass percent of magnesium is calculated as 1 mole of magnesium having 24.305 grams/mole will have weight of 24.305 grams and 1 mole of MgSO4 will have 120.366 grams
Putting the values in the equation:
24.305 ÷ 144.671 × 100
= 16.8% of magnesium is in the mixture
The mass percentage of SO4 is calculated as
= 96.06 ÷ 216.426 × 100
= 44.38 %
The mass percentage of the mixture MgSO4 is 44.38 + 16.8 = 61.8 %
Mass percentage is a representation of the concentration of element or elements in a compound.
When acetylene gas, C2H2, reacts with oxygen gas the products are carbon dioxide and water. a. What is the balanced equation? b. How many grams of water can be formed from the consumption of 8.98 grams of acetylene c. How many grams of water can be formed from the consumption of 4.58 grams of oxygen d. What is the theoretical yield of the reaction ?e. The reaction yielded 1.00g of water. What is the percent yield?
Answer:
a.
C₂H₂ (g) + 5/2 O₂ (g) ⇒ 2CO₂ (g) + H₂O (l)
b. 6.21 g H₂O
c. 1.08 g
d. 6.21 g H₂O
e. 16 %
Explanation:
This question involves a calculation based on the stoichiometry of the balanced chemical equation:
b.
Lets calculate the # moles C₂H₂ 8.98 g will represent and then calculate the amount of water produced as follows:
# moles C₂H₂ = mass/molar mass = 8.98 g / 26.04 g/mol = 0.34 mol
From the stoichiometry of the reaction:
1 mol H₂O produced / mol C₂H₂ x 0.34 mol C₂H₂ = 0.34 mol H₂O produced
g H₂O = # mol H₂O x molar mass H₂O = 0.34 mol x 18.01 g/mol = 6.21 g H₂O
c.
For 4.58 g O₂ we can calculate the amount of water in grams formed as follows:
# mol O₂ = mass / molar mass O₂ = 4.58 g / 32 g / mol = 0.14 mol
From the stoichiometry of the reaction we have
1 mol H₂O produced /2.5 mol O₂ x 0.14 mol O₂ = 0.06 mol H₂O
mass H₂O produced = 0.06 mol x molar mas H₂O = 0.06 mol x 18.01 g/mol
= 1.08 g H₂O
d,e.
We calculated in part b that we should have produced 6.21 g H₂O, therefore the percent yield is =
1 g / 6.21 g x 100 g = 16 %
Note one could argue that this theoretical yield refers to the 4.58 grams O₂ in part c. However if that were the case we will have more than 100 % yield, unless we round the numbers to give us 100 % yield
Final answer:
To find the grams of water formed from acetylene or oxygen, we use stoichiometry, converting grams to moles and using the balanced equation to relate substances. The theoretical yield is the maximum product amount if all the limiting reactant is used, and the percent yield is actual over theoretical yield times 100%.
Explanation:
The balanced chemical equation for the combustion of acetylene (C2H2) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is: 2C2H2(g) + 5O2(g) ightarrow 4CO2(g) + 2H2O(l)
To calculate the gram of water formed from the consumption of acetylene or oxygen, we'll use stoichiometry. For instance, from 8.98 grams of acetylene, you first determine the molar mass of C2H2 (26.04 g/mol) and convert the mass of C2H2 to moles. Then, you use the balanced equation to find the mol-to-mol ratio between acetylene and water, leading you to the moles of water. Finally, convert moles of water to grams using the molar mass of water (18.015 g/mol).
For the case of 4.58 grams of oxygen, you would do a similar calculation. Determine the molar mass of O2 (32.00 g/mol), convert the mass of O2 to moles, and use the stoichiometric ratios from the balanced equation to find the moles and then grams of water produced.
The theoretical yield is the maximum amount of product that could be formed from the given amounts of reactants, assuming complete conversion and that all of the limiting reactant is consumed.
The percent yield is calculated as the actual yield divided by the theoretical yield, multiplied by 100%. Therefore, to find the percent yield given an actual yield of 1.00g of water, you would need to compare it to the theoretical yield previously calculated.
Consider the titration of 50.0 mL of 1.00 M C5H5N by 0.500 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for C5H5N
The given question is incomplete. The complete question is as follows.
Consider the titration of 50.0 mL of 1.00 M [tex]C_{5}H_{5}N[/tex] by 0.500 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for [tex]C_{5}H_{5}N[/tex] = [tex]1.7 \times 10^{-9}[/tex]. Calculate the pH at the equivalence point for this titration.
Explanation:
Reaction equation of the given reaction is as follows.
[tex]C_{5}H_{5}N + H^{+} \rightarrow C_{5}H_{5}NH^{+}[/tex]
Only [tex]C_{5}H_{5}N[/tex] is present in the solution before the addition of HCl. Hence, the equation will be as follows.
[tex]C_{5}H_{5}N + H_{2}O \overset{K_{b}}{\rightleftharpoons} C_{5}H_{5}NH^{+} + OH^{-}[/tex]
And, at the equivalence point entire [tex]C_{5}H_{5}N[/tex] will completely react with the HCl. Hence, the solution contains [tex]C_{5}H_{5}NH^{+}[/tex] at the equivalence point. It is acidic with a pH less than 7.
No. of moles of HCl = No. of moles of [tex]C_{5}H_{5}N[/tex]
[tex]V_{HCl} \times M_{HCl} = V_{C_{5}H_{5}N} \times M_{C_{5}H_{5}N}[/tex]
[tex]V_{HCl} \times 0.5 M = 50 ml \times 1 M[/tex]
[tex]V_{HCl}[/tex] = 100 ml
Hence, the volume of HCl at the equivalence point is 100 ml.
Also, [tex]k_{a} = \frac{k_{w}}{k_{b}}[/tex]
= [tex]\frac{1 \times 10^{-14}}{1.7 \times 10^{-19}}[/tex]
= [tex]5.9 \times 10^{-6}[/tex]
Therefore, concentration of the acid is calculated as follows.
Concentration = [tex]\frac{\text{No. of moles of acid}}{\text{Volume of solution}}[/tex]
= [tex]\frac{0.05}{0.15}[/tex]
= 0.333 M
Since, [tex]C_{5}H_{5}NH^{+}[/tex] is a weak acid. So,
[tex][H^{+}] = \sqrt{k_{a} \times C}[/tex]
= [tex]\sqrt{5.9 \times 10^{-6} \times 0.333 M}[/tex]
= [tex]1.40 \times 10^{-3}[/tex]
Now, we will calculate the pH of the solution at the equivalence point as follows.
pH = [tex]-log (1.40 \times 10^{-3})[/tex]
= 2.85
Thus, we can conclude that pH at the equivalence point for this titration is 2.85.