1. Col2
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Answer:
The amount of NO₂ that can be produced 8.533 g
Explanation:
According to question
2 NO(g) + O₂(g) → 2 NO₂(g)
Given
Moles of nitrogen monoxide = 0.377
Moles of oxygen = 0.278
Since 'NO' is the limiting reagent according to this ratio.
According to equation
2 moles NO reacts to form 2 moles NO₂
So, 0.1855 moles NO give = 0.1855 moles of NO₂
Mass of 1 mole NO₂ = 46 g/mole
Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g
W=F*d
W= 500 J
F = 250 N
500 J = 250 N * d
d= 500J/250 N = 2 J/N = 2(N*m)/N = 2 m
Answer is 2 m.
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