Nonfoliated is the answer I belive.. Hopefully
The answer is 0.375 mol of silver chromate
The molarity of the acid given the data from the question is 0.30 M
<h3>Balanced equation </h3>
2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HNO₃ (nA) = 2
- The mole ratio of the base, Ba(NO₃)₂ (nB) = 1
<h3>How to determine the molarity of the acid</h3>
From the question given above, the following data were obtained:
- Volume of acid, HNO₃ (Va) = 39.7 mL
- Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
- Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
- Molarity of acid, HNO₃ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 39.7) / (0.25 × 24) = 2
(Ma × 39.7) / 6 = 2
Cross multiply
Ma × 39.7 = 6 × 2
Ma × 39.7 = 12
Divide both side by 39.7
Ma = 12 / 39.7
Ma = 0.30 M
Learn more about titration:
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The balanced chemical reaction is:
<span>CuCl2 + 2Na → 2NaCl + Cu
We are given the amount of sodium to be used up in the reaction. This will be the starting point for our calculations.
15 g Na ( 1 mol / 22.99 ) ( 1 mol Cul2 / 2 mol Na ) (134.45 g / 1 mol ) = 43.86 g CuCl2 needed to be able to obtain the maximum amount of copper.</span>
