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3 years ago

61.0 mol of P4O10 contains how many moles of P

Chemistry

2 answers:

galina1969 [7]3 years ago

5 0

61mol * 4 = 244moles of P

Elanso [62]3 years ago

4 0

Answer:

244 moles of phosphorus are present in 61.0 moles of $P_4O_{10}$ .

Explanation:

Moles of $P_4O_{10}$ = 61.0 moles

1 mol of $P_4O_{10}$ contains 4 moles of phosphorus.

Then 61.0 moles of $P_4O_{10}$ will contain:

$61.0 mol\times 4=244 mol$ of phosphorus

244 moles of phosphorus are present in 61.0 moles of $P_4O_{10}$ .

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PbO2 + 4HCl --- PbCl2 + Cl2 + 2H2O who buys electrons and who loses electrons?

SIZIF [17.4K]

Answer: Electrons are taken up by $PbO_2$ and they are lost by $HCl$

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is also called the reaction where the exchange of electrons takes place.

An oxidation reaction is defined as the reaction in which a chemical species loses electrons takes place. In this reaction, the oxidation state of a substance gets increased.

A reduction reaction is defined as the reaction in which a chemical species gains electrons takes place. In this reaction, the oxidation state of a substance gets reduced.

For the given chemical reaction:

$PbO_2+4HCl\rightarrow PbCl_2+Cl_2+2H_2O$

The half-reactions for this redox rection follows:

Oxidation half-reaction: $2HCl\rightarrow ClO_2 + 2e^-$

Reduction half-reaction: $PbO_2+2e^-\rightarrow PbCl_2$

Hence, electrons are taken up by $PbO_2$ and they are lost by $HCl$

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3 years ago

An element that is malleable and a good conductor of heat and electricity could have an atomic number of.....

lilavasa [31]

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3 years ago

What is the primary pigment found in the chlotopast

Lady_Fox [76]

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3 years ago

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ra1l [238]

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3 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago