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3 years ago

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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Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

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⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

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∴ [A] = 1.6 M

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∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

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3 years ago

Convert 125 g to kilograms

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3 years ago

Read 2 more answers

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

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