the cathode is where reduction occurs!
Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
<span>R = 0.08206
L·atm/mol·K.
Ideal gas law: p·V = n·R·T.</span>
V(O₂) =
n·R·T / p.<span>
V(O₂) =
0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.</span><span>
V(O₂<span>) =
4.63 L.</span></span>
There are 2071.79 grams of glucose in 11.5 moles.
By adding the enthalpies of the intermediate reactions together to get the enthalpy of the desired reaction
Answer:
Mole fraction of 
= 0.58
Mole fraction of 
= 0.42
Explanation:
Let the mass of and = x g
Molar mass of = 33.035 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molar mass of = 46.07 g/mol
Thus,
So, according to definition of mole fraction:
Mole fraction of = 1 - 0.58 = 0.42
