Answer:
0.58 atm
Explanation:
Step 1: Given data
- Total pressure of the gaseous mixture (P): 1.05 atm
- Partial pressure of N₂ (pN₂): 0.35 atm
- Partial pressure of H₂ (pH₂): 0.12 atm
- Partial pressure of CO₂ (pCO₂): ?
Step 2: Calculate the partial pressure of CO₂
The total pressure of the gaseous mixture is equal to the sum of the partial pressures of the individual gases.
P = pN₂ + pO₂ + pCO₂
pCO₂ = P - pN₂ - pO₂
pCO₂ = 1.05 atm - 0.35 atm - 0.12 atm = 0.58 atm
D. She is not right, because there will be more successful collisions
between reactants in the concentrated solutions.
Answer:
This is a simple case of ratios. (1 mol)/(22.4 L)=(n mol)/(.025 L) Then we cross multiply and we get 22.4n=.025 We divide each side by 22.4 to find n=.001116 mol Then to convert the moles to atoms we multiply, and cross-cancel the units (.001116 mol)/1 xx (6.02 xx 10^23 atms)/(1 mol) and we have 6.72 xx 10^20 atoms. I've found the trick of cross-cancelling units to be a very effective mnemonic, it always makes sure you carry out the correct calculation to find the desired units.
Explanation:
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
