Answer:
ΔG°′ = 1.737 KJ/mol
Explanation:
The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.
The overall equation of reaction is as follows:
fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V
Using the equation for standard free energy change; ΔG°′ = −nFΔE°′
where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V
ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V
ΔG°′ = 1.737 KJ/mol
Answer:
6.286 L.
Explanation:
Applying Charles Law,
V/T = V'/T'............................... Equation 1
Where V = Initial volume of the balloon, T = Initial Temperature of the balloon, V' = Final volume of the balloon, T' = Final Temperature of the balloon
make V' the subject of the equation
V' = VT'/T............................ Equation 2
Given: V = 6.50 L, T = 31 °C = (273+31) K = 304 K, T' = 21 °C = (273+21) K = 294 K
Substitute into equation 2
V' = 6.5(294)/304
V' = 6.286 L.
Hence the new volume in the balloon = 6.286 L.
To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer:
165 of CO₂.
Explanation:
In the reaction:
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
2 moles of HCl reacts producing 1 mole o CO₂
If 7.5 moles of HCl reacts, moles of CO₂ produced are:
7.5 moles of HCl ₓ ( 1 mol CO₂ / 2 mol HCl) = 3.75 mol CO₂. As molar mass of CO₂ is 44g/mol, mass of CO₂ is:
3.75 mol CO₂ ₓ (44g / 1mol) = <em>165 of CO₂ </em>
Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
- CuSO₄: 1 mole
- Zn: 1 mole
- Cu: 1 mole
- ZnSO₄: 1 mole
Being:
- Cu: 63.54 g/mole
- S: 32 g/mole
- O: 16 g/mole
- Zn: 65.37 g/mole
the molar mass of the compounds participating in the reaction is:
- CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/mole
- Zn: 65.37 g/mole
- Cu: 63.54 g/mole
- ZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/mole
Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
- CuSO₄: 1 moles* 160 g/mole= 160 g
- Zn: 1 mole* 65.37 g/mole= 65.37 g
- Cu: 1 mole* 63.54 g/mole= 63.54 g
- ZnSO₄: 1 mole* 161.37 g/mole= 161.37 g
Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
mass of Zn= 185.49 grams
<u><em>185.49 grams of Zinc would react with 454g (1lb) of copper sulfate</em></u>
