The compound Lead(II) Nitrate reacts with potassium Iodide.

Chemistry

1 answer:

marshall27 [118]4 years ago

7 0

Answer:

a) Pb(NO3)2 + 2KI ⇒ PbI2 + 2K(NO3)

ion that form the precipitate = Pb2+ and I-

Spectator ions = NO3. and K+

b) Pb 2+ + 2 I- ⇒ PbI2

Explanation:

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A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con

Nataly_w [17]

Answer:

$\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

$Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}$

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

$Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J$

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

$n=1.400g*\frac{1mol}{89.05g} =0.01572mol$

Thus, we obtain:

$\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}$