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Naddika [18.5K]
3 years ago
15

Goodyear tires are listed with 29.72 psi pressure limits in the summer, with 10.22 L of air. In the winter, however, the volume

decreases to 7.19 L because of the colder weather. What would the be the new pressure?
Chemistry
1 answer:
serg [7]3 years ago
6 0

Answer:

42.24 psi.

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 29.72 psi

Initial volume (V₁) = 10.22 L

Final volume (V₂) = 7.19 L

Final pressure (P₂) =?

The final pressure can be obtained by applying the Boyle's law equation as follow:

P₁V₁ = P₂V₂

29.72 × 10.22 = P₂ × 7.19

303.7384 = P₂ × 7.19

Divide both side by 7.19

P₂ = 303.7384 / 7.19

P₂ = 42.24 psi

Therefore, the final pressure is 42.24 psi

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Explanation:

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3 0
2 years ago
Read 2 more answers
How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --&gt; __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
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  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

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