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DaniilM [7]
3 years ago
10

A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to

Chemistry
1 answer:
Simora [160]3 years ago
8 0

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

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Q = 2.415kj

Explanation:

Q = Mc∆temperature

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mass = 150g = 0.15kg

∆temperature = 60 - 25 = 35°c

Q = 0.15 X 0.46 X 35 = 2.415kj

Q = 2.415kj

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You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam
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Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression d=\frac{m}{V} where m is the mass and V is the volume, therefore:

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d_{A}=\frac{m_{A}}{V_{A}}

- Density of the cube B:

d_{B}=\frac{m_{B}}{V_{B}}

- Density of the cube C:

d_{C}=\frac{m_{C}}{V_{C}}

Solving for mass:

m_{A}=d_{A}*V_{A}

m_{B}=d_{B}*V_{B}

m_{C}=d_{C}*V_{C}

And all the three cubes have the same mass, so:

m_{A}=m_{B}=m_{C}

Therefore:

d_{A}*V_{A}=d_{B}*V_{B} (Eq.1)

d_{A}*V_{A}=d_{C}*V_{C} (Eq.2)

Solving for d_{1} in Eq.1:

d_{A}=d_{B}\frac{V_{B}}{V_{A}}

Replacing values for the volume:

d_{A}=d_{B}\frac{16.7mL}{25.9mL}

d_{A}=d_{B}*0.64

As we know the density of the aluminum is 2.7\frac{g}{cm^{3}}, so replacing this value for d_{B}:

d_{A}=2.7\frac{g}{mL}*0.64

d_{A}=1.728\frac{g}{mL}

that is the density of the magnesium.

Solving for d_{C} in Eq.2:

d_{C}=d_{A}\frac{V_{A}}{V_{C}}

d_{C}=d_{A}\frac{25.9mL}{4.29mL}

d_{C}=d_{A}*6.04

d_{C}=1.728\frac{g}{mL}*6.04

d_{C}=10.4\frac{g}{mL}

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

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