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nadezda [96]
3 years ago
13

an empty bottle has a mass of 24,25 grams and 86,55 grams completely full-filled with water. Then, water in the bottle is emptie

d and bottle is filled again with 123,95 grams of carbon tetrachloride solvent. Which is the density of the solvent?
Physics
1 answer:
ladessa [460]3 years ago
4 0
The density of a substance is the quotient we obtain when we divide its mass by the volume. The density is,
                          density = mass / volume
The mass of carbon tetrachloride is given to be 123.95 grams and that the volume is obtained by subtracting final weight of the completely full-filled with water bottle with the initial weight.
                              86.55 - 24.25 = 62.3 grams
Since the density of water is 1 grams/ cc. Then, the volume of the bottle is also 62.3 cc. The density is therefore,
                       density = 123.95 grams/ 62.3 cc = 1.99 g/cc
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Explanation:

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3 years ago
Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C
SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

8 0
3 years ago
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3 years ago
In a tug of war, two teams exerts a force of 30 N each. What is the net force acting on the rope?
Arte-miy333 [17]

The force exerted by each team is 30 N. The forces are in the direction opposite to each other.

Force exerted by team 1, F_{1}=30 N

Force exerted by team 2,F_{2}=-30 N

Net force, F=F_{1} +F_{2}

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Therefore, net force on the rope is 0 N.

4 0
3 years ago
Someone please help me
Trava [24]

Answer:

The answer to your question is: D) Ф₂ = 49.71°

Explanation:

Data

n₁ = 1.33

Ф₁ = 35°C

n₂ = 1

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Process

Substitution

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Ф₂ = sin⁻¹ (1.33 sin 35/1)

Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)

Ф₂ = sin⁻¹ ( 0.7628 / 1)

Ф₂ = sin⁻¹ (0.7628)

Ф₂ = 49.71°

8 0
4 years ago
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