Answer:
0.5ppm
Explanation:
Step 1:
Data obtained from the question.
Volume of water = 2500L
Mas of Cu = 1.25 g
Step 2:
Determination of the concentration of Cu in g/L. This is illustrated below:
Volume of water = 2500L
Mas of Cu = 1.25 g
Conc. of Cu In g/L =?
Conc. g/L = Mass /volume
Conc. of Cu in g/L = 1.25/2500
Conc. of Cu in g/L = 5x10^–4 g/L
Step 3:
Conversion of the concentration of Cu in g/L to ppm. This is illustrated below
Recall:
1g/L = 1000mg/L
Therefore, 5x10^–4 g/L = 5x10^–4 x 1000 = 0.5mg/L
Now, we know that 1mg/L is equal to 1ppm.
Therefore, 0.5mg/L is equivalent to 0.5ppm
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.
Answer:
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