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3 years ago

How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?

Chemistry

2 answers:

damaskus [11]3 years ago

6 0

Answer:

The correct answer is option D.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 233 kPa

1 atm = 101.325 kPa

$233kPa=\frac{233}{101.325} atm=2.30 atm$

V = Volume of gas = 120 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 340 K

Putting values in above equation, we get:

$2.30 atm\times 120L=n\times (0.0821L.atm/mol.K)\times 340K\\\\n=9.88 mole\approx 9.9 mole$

natka813 [3]3 years ago

5 0

A would be correct have a nice day.

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4 years ago

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jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

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As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

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We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

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So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

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As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

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First we have to calculate the pH.

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Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

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$pOH=-\log [OH^-]$

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3 years ago

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Bad White [126]

Answer:

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