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astraxan [27]
3 years ago
13

P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

P4O10(s) + 6H2O(l) → 4H3PO4(s)
Chemistry
1 answer:
mash [69]3 years ago
6 0

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

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julia-pushkina [17]

Answer:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

HI(aq)+NaOH(aq)\rightarrow NaI+H_2O

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

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Which is also balanced as the number of atoms of all the elements is the same at both sides.

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