Answer:
334J/g
Explanation:
Data obtained from the question include:
Mass (m) = 1g
Specific heat of Fusion (Hf) = 334 J/g
Heat (Q) =?
Using the equation Q = m·Hf, we can obtain the heat released as follow:
Q = m·Hf
Q = 1 x 334
Q = 334J
Therefore, the amount of heat released is 334J
The answer is ................................ c
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
