Answer:
(1) I shifts toward product and II shifts toward reactant.
Explanation:
Increasing the temperature of an endothermic reaction (∆H is positive) shifts the equilibrium position to the right thus favoring product formation.
Increasing the temperature of an exothermic reaction (∆H is negative) shifts the equilibrium position to the left thus favoring the backward reaction.
Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
2SO2 + O2 ------> 2SO3
1) M(SO2)= 32.0 + 2*16.0 = 64 g/mol
2) 100.0 g SO2 * 1 mol SO2/64 g SO2 = 1.5625 mol SO2
3) 2SO2 + O2 ------> 2SO3
2 mol 1 mol
1.5625 mol x mol
x= 1.5625/2=0.78125 ≈ 0.7813 mol O2
Answer: 0.7813 mol O2.
Hmm, Sounds like the best answer would be Surface Bound Monomers. Hope this was correct :)
Answer:
3 holes
red ball
hard shiny surface
Explanation:
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