The atomic mass of the metal.(M) = 51 g/mol
<h3>Further explanation</h3>
Given
32% of oxygen
Required
the atomic mass of the metal.
Solution
An oxide of trivalent metal : M₂O₃(Ar O = 16 g/mol)
The molar mass of metal oxide = 2M+16.3=2M+48
32% of oxygen, then :
32% x 2M+48 = 48
0.32(2M+48)=48
0.64M+15.36=48
0.64M=32.64
M=51
1. high
2. roads
3. improved
i’m assuming these are the answers, but fhwjfjwjf its hard to tell when it’s not multiple choice
Explanation:
First Question:
The equation is given as;
2KClO3 --> 2KCl + 3O2
2 mol of KClO3 produces 2 mol of KCl
Converting 7.6 g of KCl to mol;
Number of moles = Mass / Molar mass = 7.6 / 74.5513 = 0.1019
2 mol = 2 mol
x mol = 0.1019
solving for x;
x = 0.1019 mol of KClO3
converting to mass;
Mass = Number of mol * Molar mass = 0.1019 * 122.55 = 12.48
Correct option = B. 12.5 g
Second Question
The equation is given as;
1N2 + 3H2 --> 2NH3
3 mol of H2 produces 2 mol of NH3
Converting 34.06g of NH3 to mol;
Number of moles = Mass / Molar mass = 34.06 / 17.031 = 2 mol
3 mol = 2 mol
x mol = 2 mol
Solving for x;
x = 6 / 2 = 3 mol of H2
Converting 3 mol H2 to g;
Mass = Number of moles * Molar mass = 3 * 2.016 = 6.048g
Correct option: B. 6.05g
You can calculate the approximate final volume using the ideal gas law at constant pressure.
This is the reknown Law of Charles:
V / T = constant.
=> V1 / T1 = V2 / T2
Where V is the volume and T is the absolute temperature (Kelvin).
V2 = (V1 / T1) * T2
V1 = 4.1 liter
T1 = 22 + 273.15 K = 295.15K
T2 = 52 + 273.15 K = 325.15K
V2 = (4.1 liter / 295.15K)*325.15K = 4.6 liter
Answer: 4.6 liter
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