55.0 mL of 0.200 M H3PO4 is mixed with 55.0 mL of 0.600 M KOH intially at 21.62C Predict the final temperature of the solution if its density is 1.13 g/ml and its specfic heat is 3.78J/(gC) Assume the total volcume is the sume of the individual volumes
h3 po4(aq)+3koh(aq) to 3h2o(l)+k3po4(aq)+173.2
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Expert's answer
2018-03-07T13:20:06-0500
H3PO4 solution contains: 55.0 mL x 0.200 M / 1000 mL = 0.011 mol of acid KOH solution contains: 55.0 mL x 0.600 M / 1000 mL = 0.033 mol af base (3 times more that the acid, the nessesary quantity to react completely)
According to the reaction equilibrium: 1 mole of acid reacts and produses 173.2 kJ 0.011 moles – x X = 1.9052 kJ = 1905.2 J.
Q = cm(T2 – T1)
The mass of final sln: 55 mL + 55 mL = 110 mL 110 mL x 1.13 = 124.3 g
1905.2 J = 3.78 J/gC x 124.3 g (T2 - 21.62) T2 = 25.67 C Answer: 25.67 C.
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07.03.18, 19:35
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lj
07.03.18, 14:56
0.260 mol of HI gas is placed in a 2.00 L container at 450 degree
Celsius and allowed to equilibrate by the reaction 2HI(g)= h2(g) + i2
(g), at equilibrium the concentration of i2 gas is 0.0150M, what is
the value of k for this reaction at this temperature?
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Dear visitor, please use panel for submitting new questions
0.260 mol of HI gas is placed in a 2.00 L container at 450 degree Celsius and allowed to equilibrate by the reaction 2HI(g)= h2(g) + i2 (g), at equilibrium the concentration of i2 gas is 0.0150M, what is the value of k for this reaction at this temperature?
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