Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
The correct choice in the options above is the aerosols. It is because the aerosols are the ones that are combined with gaseous substances and water in order for it to be formed. Without the gaseous substance being joined with the water then the aerosols won't be produced.
Answer:
a) rate law1 = k[NO2]²
b) rate law2 = k[NO][O3]
Explanation:
NO2(g) + CO(g) → NO(g) + CO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
When [NO2] in reaction 1 is doubled, the reaction quadruples
Rxn is second order.
rate law1= [NO2]^a [CO]^b
rate law1= [NO2]² [CO]^0
rate law1 = k[NO2]²
When [NO] in reaction 2 is doubled, the rate doubles.
Rxn is first order
The ratio is 1:1
this makes the rate law2 = k[NO][O3]
Answer:
8.61 mL of the HCl solution
Explanation:
The reaction that takes place is:
- 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
From the given mass of Mg(OH)₂, we can calculate <u>the moles of HCl that are neutralized</u>:
- 4x10² mg = 400 mg = 0.400g
- 0.400g Mg(OH)₂ ÷ 58.32g/1mol = 6.859*10⁻³ mol Mg(OH)₂
- 6.859*10⁻³ mol Mg(OH)₂ *
3.429x10⁻³ mol HCl
Finally, to calculate the volume of an HCl solution, we need both the moles and the concentration. We can <u>calculate the concentration using the pH value</u>:
= [H⁺]
- 0.0398 M = [H⁺] = [HCl] *Because HCl is a strong acid*
Thus, the volume is:
- 0.0398 M = 3.429x10⁻³mol HCl / Volume
- Volume = 8.616x10⁻³ L = 8.62 mL
Answer:
0.558 M
Explanation:
Data obtained from the question. This includes the following:
Initial concentration (C1) =..?
Initial volume (V1) = 25mL
Final volume = 25 + 50 = 75mL
Final concentration (C2) = 0.186 M
The initial concentration of the solution can be obtained as follow:
C1V1 = C2V2
C1 x 25 = 0.186 x 75
Divide both side by 25
C1 = (0.186 x 75) /25
C1 = 0.558 M
Therefore, the initial concentration of the stock was 0.558 M
