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34kurt
2 years ago
7

A sample of an unknown gas occupies 5.51 L at 1.31 atm. What pressure would thisgas exert in a 0.520 L container if the temperat

ure is held constant?
Chemistry
1 answer:
weeeeeb [17]2 years ago
5 0

Answer:

P₂ = 13.9 atm (3 sig. figs.)

Explanation:

The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...

P ∝ 1/V => P = k/V => k = P·V

For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂

Given:

P₁ = 1.31 atm.

V₁ = 5.51 L

P₂ = ?

V₂ = 0.520 L

V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)

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|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

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|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

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The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

Cu (OH)2 =  \frac{1.8 \times  {10}^{ - 7} mol \:Cu (OH)2 }{1L}  \times  \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2}  \\  = 1.75428 \times 10 ^{ - 5}

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

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