<u>Answer:</u> The partial pressure of hydrogen is 93.9 kPa.
<u>Explanation:</u>
To calculate the partial pressure of hydrogen, we will follow Dalton's Law.
This law states that the total pressure of a mixture of gases is equal to the sum of the individual pressures exerted by the constituent gases.
Mathematically,
According to the question,
We are given:
Putting values in above equation, we get:
Hence, the partial pressure of hydrogen is 93.9 kPa.
Answer:
The answer is 0.36 kg/s NO
Explanation:
the chemical reaction of NH3 to NO is as follows:
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
We have the following data:
O2 Volume rate = 645 L/s
P = 0.88 atm
T = 195°C + 273 = 468 K
NO molecular weight = 30.01 g/mol
we calculate the moles found in 645 L of O2:
P*V = n*R*T
n = P*V/R*T
n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2
With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:
14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO
Answer:
b
Explanation:
orbital overlap electron densities.
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
B. The limiting reactant determines the max amount of product that can be formed
