<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
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Answer:
1
Explanation:
4 HBr + O2 → 2H 20 + 2Br 2
...............
Mass = mr x moles
Mr of CuCl2 = ( 63.5) + ( 35.5 x 2) = 134.5
2.5 = 134.5 x moles
2.5 / 134.5 = moles
Moles = 0.019 (2DP)
0.25g of Al
Mr of Al = 27
0.25 = 27 x moles
0.25/ 27 = 0.0093 moles (2sf)
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