X is always the independent variable
The balanced chemical reaction:
K2SO4 + O2 = 2KO2 + SO2
Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:
7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)

R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
Answer is: f<span>ormula for the hydrated compound is CuSO</span>₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
0,253 : M(H₂O) = 0,747 : 159,6 g/mol.
M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.