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3 years ago

6

A 55.0 L sample of gas at 235 K exerts a pressure of 225 KPa. If its volume is decreased to 42.0 L and the temperature rises to

315 K, what is its new pressure?

1 answer:

OleMash [197]3 years ago

4 0

Answer:

Final pressure will be equal to 394.94 kPa

Explanation:

It is given volume $V_1=55L$

Temperature of the gas $T_1=235K$

Initial pressure exerted $P_1=225kPa$

Now volume is decreased to $V_2=42L$

And temperature is increases to $T_2=315K$

By using gas equation

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\frac{225\times 55}{235}=\frac{P_2\times 42}{315}$

$P_2=394.94kPa$

So final pressure will be equal to 394.94 kPa

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Explanation :

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2 years ago

1. How many molecules are in 240g of sulfuric acid?

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14.8 × 10²³ molecules

Explanation:

Given data:

Mass of sulfuric acid = 240 g

Number of molecules = ?

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The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

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<em>Number of moles = mass/ molar mass</em>

Number of moles = 240 g/ 98 g/mol

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1 mole = 6.022 × 10²³ molecules

2.45 × 6.022 × 10²³ molecules

14.8 × 10²³ molecules

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3 years ago

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The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

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