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kherson [118]
3 years ago
5

What is the volume, in liters, of 2.00 moles of hydrogen at STP?

Chemistry
1 answer:
Ber [7]3 years ago
8 0
The volume of one mole of any gas at STP is 22.4 L. So, at STP, the volume of 2.00 moles of hydrogen gas would be (22.4 L/mol)(2 mol H2) = 44.8 L.
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What is the balance equation for HgO(s)- Hg(l)+O2(g)
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The balanced equation is   2HgO --> 2Hg + O2
4 0
3 years ago
Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i
marta [7]
When you want to melt an ice, you only need the latent energy of fusion, <span>δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

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3 0
3 years ago
Volume of 8.29ml and a mass of 16.31g
strojnjashka [21]
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Hope this helps!
3 0
3 years ago
Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at
Svetllana [295]

Answer:

The rate at which P_4 is being produced is 0.0228 M/s.

The rate at which PH_3 is being consumed is 0.0912 M/s.

Explanation:

4PH_3\rightarrow P_4(g)+6H_2(g)

Rate of the reaction : R

R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which hydrogen is being formed = \frac{d[H_2]}{dt}=0.137 M/s

R=\frac{1}{6}\frac{d[H_2]}{dt}

R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s

The rate at which P_4 is being produced:

R=\frac{1}{1}\frac{d[P_4]}{dt}

0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which PH_3 is being consumed :

R=\frac{-1}{4}\frac{d[PH_3]}{dt}

0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}

\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s

6 0
3 years ago
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