<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>
They will most likey group with the second group. Since they have 6 electrons and want to have a full outer shell so therefore would group with the second group.
Answer:
controlled by the Arabs, who brought frankincense and myrrh by camel caravan from South Arabia.
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:
E = m(ΔH)
where:
m = mass of substance
ΔH = latent heat of fusion or melting
According to data, the ΔH of mercury is approximately 11.6 kJ/kg.
E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J
The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
