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Anna71 [15]
4 years ago
7

15 POINTS FOR ANYONE WHO CAN ANSWER THIS!!!!!! plZ

Chemistry
1 answer:
yaroslaw [1]4 years ago
8 0

Answer:

A. 2Al(s) + 3Cl2(g) —> 2AlCl3(s)

B. 2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

C. H2(g) + Br2(l) —> 2HBr(g)

Explanation:

A. Reaction between aluminium metal and chlorine gas.

Al(s) + Cl2(g) —> AlCl3(s)

The above equation can be balance as follow:

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of AlCl3 as shown below:

Al(s) + 3Cl2(g) —> 2AlCl3(s)

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al(s) + 3Cl2(g) —> 2AlCl3(s)

Now the equation is balanced.

B. Reaction between lithium metal and liquid water.

Li(s) + H2O(l) —> LiOH(aq) + H2(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and a total of 3 atoms on the right side. It can be balance by putting 2 in front of H2O and 2 in front of LiOH as shown below:

Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

There are 2 atoms of Li on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Li as shown below:

2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

Now, the equation is balanced.

C. Reaction between gaseous hydrogen and liquid bromine.

H2(g) + Br2(l) —> HBr(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and 1 atom on the right. It can be balance by putting 2 in front of HBr as shown below:

H2(g) + Br2(l) —> 2HBr(g)

Now the equation is balanced.

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A mixture of H2 and water vapor is present in a closed vessel at 20. 00°C. The total pressure of the system is 755. 0 mmHg. The
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The partial stress of H2 is 737.47 mmHg Let's observe the Ideal Gas Law to find out the whole mols.

We count on that the closed vessel has 1L of volume

  • P.V=n.R.T
  • We must convert mmHg to atm. 760 mmHg.
  • 1 atm
  • 755 mmHg (755/760) = 0.993 atm
  • 0.993 m.1L=n.0.082 L.atm/mol.K .
  • 293 K(0.993 atm 1.1L)/(0.082mol.K /L.atm).
  • 293K = n
  • 0.0413mols = n

These are the whole moles. Now we are able to know the moles of water vapor, to discover the molar fraction of it.

  1. P.V=n.R.T
  2. 760 mmHg. 1 atm
  3. 17.5 mmHg (17.5 mmHg / 760 mmHg)=0.0230 atm
  4. 0.0230 m.1L=n.0.082 L.atm/mol.K.293 K(0.0230atm.1L)/(0.082mol.K/L.atm .293K)=n 9.58 × 10 ^ 4 mols = n.
  5. Molar fraction = mols )f gas/general mols.
  6. Molar fraction water vapor =9.58×10^ -four mols / 0.0413 mols
  7. Sum of molar fraction =1
  8. 1 - 9.58 × 10 ^ 4 × mols / 0.0413 ×mols = molar fraction H2
  9. 0.9767 = molar fraction H2
  10. H2 pressure / Total pressure =molar fraction H2
  11. H2 pressure / 55mmHg = =0.9767 0.9767 = h2 pressure =755 mmHg.
  12. 737,47 mmHg.
<h3>What is a mole fraction?</h3>

Mole fraction is a unit of concentration, described to be identical to the variety of moles of an issue divided through the whole variety of moles of a solution. Because it's miles a ratio, mole fraction is a unitless expression.

Thus it is clear that the partial pressure of H2 is 737,47 mmHg.

To learn  more about partial pressure refer to the link :

brainly.com/question/19813237

<h3 />

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