Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
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CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
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1) Ca(OH)2 should be reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s)
we are going to take as
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>
</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)
Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
Answer:
Decreasing the temperature will shift the equilibrium leftwards towards reactants.
Explanation:
Hello!
In this case, since the reaction between chromate anions and hydrogen ions yields dichromate anions, water and heat, we can infer this is an exothermic reaction by which heat is released (remember in endothermic reactions heat is absorbed as a reactant), it means that considering the LeChatelier’s which states that increasing the temperature of an exothermic reaction shifts the equilibrium leftwards since heat is a product, otherwise (decreasing the temperature) the equilibrium will be shifted rightwards.
Therefore, decreasing the temperature is the perturbation that will shift the equilibrium leftwards towards the reactants.
Best regards!
<span> ca. 0.4 moles. if this helps plz medal!</span>
Answer:
Suppose you added some solid NaCl to a saturated solution of NaCl at 20℃ and warmed the mixture to 40℃. What would happen to the added NaCl?
Explanation:
can you help with this one
the number of protons and the number of neutrons determine an element's mass number. :D
