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Anuta_ua [19.1K]
3 years ago
13

If a gas is moved from a large container to a small container but its temperature and number of moles remain the same, what woul

d happen to the pressure of the gas?
It would increase
It would be halved
It would stay the same
It would slightly decrease
Chemistry
2 answers:
bazaltina [42]3 years ago
8 0

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

<span>P2 = P1V1/V2</span>

<span>
</span>

<span>The correct answer is the first option. Pressure would increase. This can be seen from the equation above where V2 is indirectly proportional to P2.</span>

alex41 [277]3 years ago
4 0

Answer:

It would increase.

Explanation:

If the gas is moving from a larger container to a small container, the volume of the gas will decrease because the gas molecules fulfill the volume of the container they are.

With no change at the temperature and the number of moles, the process can be studied by Boyle's equation:

P1*V1 = P2*V2

Where P is the pressure, V is the volume, 1 represents the first container, and 2 the second container.

We can observe that, if the volume decreases, the pressure must increase, so equality will be true. Pressure and volume are indirectly proportional.

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What type of rock is a obsidian , marble, and a conglomentary rock
laiz [17]

Hi There!

Obsidian - Igneous Rock

Marble - Metamorphic Rock

Conglomentary - Sedimentary Rock

Hope This Helps :)

6 0
4 years ago
Read 2 more answers
How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 43.6 grams of sodium hydroxide in
GaryK [48]

Answer is: 25.84  milliliters of sodium metal.

Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.

d(Na) = 0.97 g/mL; density of sodim.

m(NaOH) = 43.6 g; mass of sodium hydroxide.

n(NaOH) = m(NaOH) ÷ M(NaOH).

n(NaOH) = 43.6 g ÷ 40 g/mol.

n(NaOH) =1.09 mol; amount of sodium hydroxide.

From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).

n(Na) = 1.09 mol.

m(Na) = 1.09 mol · 23 g/mol.

m(Na) = 25.07 g; mass of sodium.

V(Na) = m(Na) ÷ d(Na).

V(Na) = 25.07 g ÷ 0.97 g/mL.

V(Na) = 25.84 mL.

7 0
4 years ago
The emission spectrum of cesium contains two lines whose frequencies are (a)
Lady bird [3.3K]

The lines are violet and blue respectively.

<h3>What is the energy?</h3>

We know that the energy of the photon could be obtained by the use of the equation;

E = hf

E = energy

h = Plank's constant

f = frequency

For the first line;

E = 6.6 * 10^-34 Js * 3.45 x 10^14 Hz = 2.3 * 10^-19 J

Given that;

E = hc/λ

λ = hc/E

λ = 6.6 * 10^-34 * 3 * 10^8/2.3 * 10^-19

λ = 8.61 * 10^-7 m or 861 nm

The color is violet

For the second line;

E = 6.6 * 10^-34 Js * 6.53 xx 10^14 Hz

E = 4.3 * 10^-19 J

E = hc/λ

λ = hc/E

λ =  6.6 * 10^-34 * 3 * 10^8/4.3 * 10^-19

λ = 4.60 * 10^- 7 m or 460 nm

The color is blue

Learn more about emission spectrum:brainly.com/question/13537021

#SPJ1

8 0
2 years ago
Which element would not have similar properties with Calcium?
elena-14-01-66 [18.8K]
Potassium would not have similar properties because it is the alkali metals, not alkali earth metals
5 0
4 years ago
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Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
iren2701 [21]

Answer:

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb(s)  →   Pb2+(aq) + 2 e-       ε⁰ oxidation       =  -   ε⁰  reduction

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reduction    

Br2(l) + 2 e- → 2 Br-(aq)           ε⁰  reduction     = +1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction = + 0.13 V + 1.07 V  = 1.20 V

4 0
3 years ago
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