Answer:
44.8 L of N₂.
Explanation:
We'll begin by writing the balanced equation for the. This is illustrated below:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
1 mole of N₂ reacted to produce 2 moles of NH₃.
Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e
Xmol of N₂ = (1 × 4)/2
Xmol of N₂ = 2 moles
Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.
Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:
1 mole of N₂ occupies 22.4 L at STP.
Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L
Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.
Explanation:
CH3CO2H + NH3 → CH3 CO2− + NH4+
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Answer is: there are 3·10²³ atoms of carbon-12.
m(C) = 6.00 g; mass of carbon.
M(C) = 12 g/mol.
n(C) = m(C) ÷ M(C).
n(C) = 6 g ÷ 12 g/mol.
n(C) = 0.5 mol.
N(C) = Na · n(C).
n(C) = 6·10²³ 1/mol · 0.5 mol.
n(C) = 3·10²³.
n - amount of substance.
M - molar mass.
Na - Avogadro number.