Which of the following actions can a person take in order to best conserve fossil fuel resources?

How do evaporation and condensation relate to heat

ale4655 [162]

Evaporation and condensation is caused by heat. That is how they are related. Heat causes the liquid to evaporate and float into the atmosphere which causes the formation of clouds.

Hope this helped. Have a great day!

7 0

3 years ago

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A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon

Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0

2 years ago

Chemistry. hellppp. pls

AveGali [126]

Carbon dioxide consists of the elements Carbon (C) and Oxygen (O) and you have to have 2 oxygen.

6 0

2 years ago

The chemical equation below is correctly balanced.

DIA [1.3K]

Answer:

44.8 L of O2 will react (option D)

Explanation:

Step 1: Data given

Number of moles of SO2 = 4.00 moles

STP = Pressure = 1 atm and temperature = 273 K

Step 2: The balanced equation

2 SO2(g) + O2(g) → 2 SO3(g)

Step 3: Calculate moles of O2

For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3

For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2

Step 4: Calculate volume of O2

For 1 mol we have a volume of 22.4 L

V = (n*R*T)/ p

V = (2.00 * 0.08206 * 273)/p

V = 44.8 L

For 2.00 moles we have a volume of 2*22.4 = 44.8 L

44.8 L of O2 will react (option D)

8 0

3 years ago

A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

Alja [10]

Answer:

Question 1: 0.2J/(gºC)

Question 2: 6,000J

Question 3: 300J

Question 4: 80g

Question 5: 74ºC

Question 6: 50g

Explanation:

Question 1. A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.

The thermal energy equation is:

Q = m × C × ΔT

Substitute and solve for C:

566J = 20g × C × (195ºC - 35ºC)

C = 566J / (20g × 160ºC)

C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

You must round to one significant figure because one factor has one significant figure).

Qustion 2. 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)

Use the thermal energy equation again:

Q = m × C × ΔT

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

Question 3. Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

Q = 315J

Round to one significant figure (because 50g has one significant figure)

Q = 300J

Question 4. Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

m = 3,000J / (0.712J/gºC × 50ºC)

m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.

Q = m × C × ΔT

Q is negative, since it is released.

Substitute and solve for T:

- 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

T = - 7500J / 400g × 4.184J/gºC) + 70ºC

T = 74ºC

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

Question 6. How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC

Q = m × C × ΔT

Subsitute:

9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

m = 50g.

8 0

2 years ago