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3 years ago

7

A sample of xenon gas at a pressure of 1.14 atm and a temperature of 20.4 °C, occupies a volume of 16.9 liters. If the gas is al

lowed to expand at constant temperature to a volume of 27.7 liters, the pressure of the gas sample will be atm.

1 answer:

kondor19780726 [428]3 years ago

8 0

Answer:

P2= 0.696atm

Explanation:

Applying Boyle's law

P1V1= P2V2

1.14×16.9= P2× 27.7

Simplify

P2= (1.14×16.9)/27.7

P2= 0.696atm

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How does your pattern of breathing change when you exercise and then rest ?

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3 years ago

Which sample of matter is classified as a solution? 1. H2O(s) 2. H2O(l) 3. CO2(g) 4. CO2(aq)

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Answer : Option 4) $CO_{2}_{(aq)}$

Explanation : $CO_{2}_{(aq)}$ is the only sample of matter which can be classified as a solution. As the solution can be defined as a liquid mixture which contains a minor component (the solute) that is uniformly distributed within the major component (the solvent).

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Therefore, $CO_{2}_{(aq)}$ is the correct answer.

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3 years ago

Read 2 more answers

For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration

Illusion [34]

Answer:

the original concentration of A = 0.0817092 M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

$k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}$

where;

k = the specific rate coefficient = 3.4 × 10⁻⁴ s⁻¹

t = time = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

$3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}$

$3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}$

$\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}= log \dfrac{a}{0.00018}$

$2.657= log \dfrac{a}{0.00018}$

$10^{2.657}= \dfrac{a}{0.00018}$

$453.94 = \dfrac{a}{0.00018}$

$a =453.94 \times 0.00018$

a = 0.0817092 M

Thus , the original concentration of A = 0.0817092 M

8 0

3 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago