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3 months ago

The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?

Chemistry

1 answer:

castortr0y [3K]3 months ago

7 0

Greetings!

The result is:

The new volume is: $2L$

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

$V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm$

Let's put the values into the equation:

$2.5L*6.0atm=7.5atm*V_{2}$

$2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L$

Consequently, the new volume is: $2L$

Wishing you a lovely day!

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When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole o

Alekssandra [3086]

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

$\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}$

The molar mass of Zn is 65 g/mole

$\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole$

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

According to the balanced reaction we find that

1 mole of $Zn$ reacts with 2 moles of $HCl$

Thus, 0.0206 moles of $Zn$ react with $0.0206\times 2=0.0412$ moles of $HCl$

This leads us to determine that $HCl$ is the excess reagent because the moles provided exceed the required moles, while $Zn$ is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat = $\frac{3.14kJ}{0.0206mol}=152.4kJ/mol$

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.

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3 months ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

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Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

According to the De-Broglie equation $\lambda = \frac{h}{p}$

with p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$ = $3.6 \times 10^{-26} J/m$

Squaring both sides gives us:

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

$12.96 \times 10^{-52}$ = $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where m = mass of the electron

Therefore, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is $7.1 \times 10^{-22} J$ .

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2 months ago

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