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Dennis_Churaev [7]
3 years ago
15

Tetraphosphorous decoxice (P4O10) reacts with water to produce phosphoric acid.

Chemistry
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

1) P4O10 + 6H20 → 4H3PO4

2)  1.25 moles of P4O10

3)  7.5 moles of water

Explanation:

1. Write the balanced equation for this reaction

P4O10 + 6H20 → 4H3PO4

2) Determine the number of moles of P4O10 required to produce 5.00 moles of phosphoric acid.

For 1 mole P4O10 consumed, we need 6 moles of water to produce 4 moles of H3PO4

If we get 5.00 moles of H3PO4, we'll have 5/4 = 1.25 moles of P4O10 consumed.

3) Determine the number of moles of water required to produce 5.00 moles of phosphoric acid

For 1 mole P4O10 consumed, we need 6 moles of water to produce 4 moles of H3PO4

If we have 5.00 moles of H3PO4, we'll get (6/4)*5 = 7.5 moles of water consumed.

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Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

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∴ Total Δs = ΔSc + ΔSh

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Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

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<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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