Asked by ash
A quantity of 16.13 mL of a KOH solution is needed to neutralize 0.4883 g of KHP. What is the concentration (in molarity) of the KOH solution?
I don't even know where to begin with the first one.
Calculate the amount of heat liberated (in kJ) from 375 g of mercury when it cools from 71.3°C to 18.0°C.
Is heat liberate a term for something else?
I don't even know where to begin with the first one.
Calculate the amount of heat liberated (in kJ) from 375 g of mercury when it cools from 71.3°C to 18.0°C.
Is heat liberate a term for something else?
Answers
Answered by
DrBob222
Convert 0.4883 g KHP to moles; moles = grams/molar mass.
Since the neutralization of KOH with KHP is 1:1 in the balanced, then moles KOH used = moles KHP.
moles KOH = M x L.
You know moles, you know L, calculate M.
For #2, yes, liberating heat means the reaction is exothermic (and delta H for the reaction is negative).
Since the neutralization of KOH with KHP is 1:1 in the balanced, then moles KOH used = moles KHP.
moles KOH = M x L.
You know moles, you know L, calculate M.
For #2, yes, liberating heat means the reaction is exothermic (and delta H for the reaction is negative).
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