In chemistry lab, a student measured the density of liquid ethanol is 0.789 g/mL. Represent its density in units of 1b /in3? (2.54 cm = 1 in., 2.205 lb = 1 kg)
A. 9.09 x 10+ 1b / in
B.4.42 x 10-1b / in?
OC.2.85 x 10-2 16 / in
D. 0.106 16 / in?
O E.5.86 x 10" 16/in3​

Answers

Answer 1

Answer:

Option C is correct option = 2.85×10⁻² lb/in³

Explanation:

Given data:

Density in g/mL = 0.789

Density in lb/in³ = ?

Solution:

It is given that,

2.54 cm = 1 in

2.205 lb = 1 kg   thus,

1 mL = 1 cm³

0.789 g/cm³ × 1 kg/ 1000 g × 2.205 lb/1 kg × (2.54 cm / 1in)³

2.85×10⁻² lb/in³

Answer 2

The density of liquid ethanol is 0.789 g/mL, which is equivalent to 0.0285 lb/in³.

A student measured the density of liquid ethanol to be 0.789 g/mL and we want to convert it to lb/in³. We will need a series of conversion factors.

What is a conversion factor?

A conversion factor is an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

Step 1: Convert 0.789 g/mL to lb/mL

We will use the following conversion factors:

1 kg = 1000 g.1 kg = 2.205 lb.

0.789 g/mL × (1 kg/1000 g) × (2.205 lb/1 kg) = 1.74 × 10⁻³ lb/mL

Step 2: Convert 1.74 × 10⁻³ lb/mL to lb/in³

We will use the following conversion factors:

1 mL = 1 cm³.1 in = 2.54 cm.

1.74 × 10⁻³ lb/mL × (1 mL/1 cm³) × (2.54 cm/1 in)³ = 0.0285 lb/in³

The density of liquid ethanol is 0.789 g/mL, which is equivalent to 0.0285 lb/in³.

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75 g of a metal is heated to a temperature of 99C. The metal is then placed in a calorimeter containing 145 g of water at a temperature of 25C. The temperature of the water in the calorimeter increase to a final temperature of 28C. What is the specific heat of the metal?

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Explanation:

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Answers

Answer:

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Explanation:

Given data:

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Percent yield = ?

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Number of moles of sodium:

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                       Na               :              H₂

                        2                :               1

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A chemist is asked to determine the specific heat capacity of an unknown mineral. The 149-g sample was heated to 92.7°C and placed into a calorimeter containing 81.4 g of water at 20.0°C. The heat capacity of the calorimeter was 12.8 J/K. The final temperature in the calorimeter was 23.7°C. What is the specific heat capacity (in J/g°C) of the mineral? Enter to 4 decimal places.​

Answers

Answer:

The specific heat of the mineral is 0.1272J/g°C

Explanation:

The sample is given energy to the calorimeter and the sample of water.

The energy released for the sample is equal to the energy absorbed for both the calorimeter and the water:

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

Where C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

The specific heat of the mineral is 0.1272J/g°C

The specific heat of the mineral is 0.1272J/g°C

Calculation of the specific heat:

The energy released for the sample should be equivalent to the energy absorbed for both the calorimeter and the water:

So,

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

here C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

Now

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

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