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?

Answers

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Explanation:

Let the specific heat of metal be s .

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= 75 x s x ( 99 - 28 ) = 5325 s

heat gained by water = mass x specific heat x gain of temperature

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Answers

Answer:

Theoretical yield = 0.1 g

Percent yield = 80%

Explanation:

Given data:

Mass of sodium = 2.3 g

Actual yield of hydrogen = 0.080 g

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Percent yield = ?

Solution:

Chemical equation:

2Na + 2H₂O    →  H₂ + 2NaOH

Number of moles of sodium:

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Now we will compare the moles of sodium with hydrogen to calculate the theoretical yield.

                       Na               :              H₂

                        2                :               1

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Mass of hydrogen produced:

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Explanation:

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Answers

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Answers

Answer:

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Molar Mass of H - 1.01 g/mol

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Step 2: Find amounts of molar masses

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1 mol Br - 79.90 g/mol

Step 3: Find molar mass of compound

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Answers

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Answers

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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

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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

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

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

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

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C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

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Answers

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Answer:

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Ionic compounds are chemical compounds that are formed by the electrostatic attraction between oppositely charged ions.

Compounds with metallic bonds and ionic compounds are similar in several ways. Some of the similarities are:

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2. Conductivity: Both metallic and ionic compounds are good conductors of electricity due to the presence of charged particles that can move freely.

3. Brittle: Ionic and metallic compounds are generally brittle in nature, meaning that they are prone to breaking or shattering when subjected to stress.

4. Solubility in polar solvents: Both metallic and ionic compounds are generally soluble in polar solvents such as water, due to the polar nature of the molecules.

In conclusion, compounds with metallic bonds and ionic compounds share several similarities. Both have high melting and boiling points, are good conductors of electricity, are brittle, and are generally soluble in polar solvents.

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