The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.
Answer is: 230 g.
Chemical reaction: P₄ + 5O₂ → 2P₂O₅.
m(P₄) = 100 g.
M(P₄) = 4 · 31 g/mol = 124 g/mol.
n(P₄) = m(P₄) ÷ M(P₄) = 100g ÷ 124g/mol = 0,806 mol.
From reaction: n(P₄) : n(P₂O5) = 1 : 2.
n(P₂O₅) = 1,612 mol.
m(P₂O₅) = 1,612 mol · 142g/mol = 230g.
M - molar mass.
n - amount of substance.
Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.
Answer:
23.71J is the work that the gas do.
Explanation:
The work that a gas do under isobaric conditions follows the formula:
W = P*ΔV
<em>Where W is work in atmL, P is the pressure and ΔV is final volume -Initial volume In Liters</em>
Replacing with the values of the problem:
W = P*ΔV
W = 0.600atm*(0.44000L - 0.0500L)
W = 0.234atmL
In Joules (1atmL = 101.325J):
0.234atmL × (101.325J / 1 atmL) =
<h3>23.71J is the work that the gas do.</h3>
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