This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
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Answer:
A substance that can conduct electricity in solution
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
3 Sig FIgs in 303,000 mm
Any nonzero is a sig fig.
Any zero between two non zeros is a sig fig.
Answer:
2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.
Explanation:
The reaction:
<em>2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.</em>
2.0 moles of hydrogen gas react with 1.0 mole of oxygen gas to produce 2.0 moles of water vapor in presence of Pt as a catalyst.
