Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
Answer:
Mass = 40.4 g
Explanation:
Given data:
Mass in gram = ?
Volume of SO₂ = 14.2 L
Temperature = standard = 273 K
Pressure = standard = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 14.2 L = n × 0.0821 atm.L/ mol.K × 273 K
14.2 atm.L = n × 22.41 atm.L/ mol
n = 14.2 atm.L/22.41 atm.L/ mol
n = 0.63 mol
Mass of sulfur dioxide:
Mass = number of moles × molar mass
Mass = 0.63 mol × 64.1 g/mol
Mass = 40.4 g
P1/V1=P2/V2
1/22.4=4/x
X=4 multiple by 22.4
V2=89.6L
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
Answer:
The mass of oxygen is 12.10 g.
Explanation:
The decomposition reaction of potassium chlorate is the following:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
We need to find the number of moles of KClO₃:
Where:
m: is the mass = 30.86 g
M: is the molar mass = 122.55 g/mol
Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3
Finally, the mass of O₂ is:
Therefore, the mass of oxygen is 12.10 g.
I hope it helps you!
