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0.100 Mole Of Lithium Has A Mass Of

Dwayne Morise

Question: 2.89 X 10²⁴

Refer to how to set up this equation in your notes.

Answer: 0.400 mole of sucrose, C₁₂H₂₂O₁₁, contains ________ C ATOMS.

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Question: 4.8 Moles

Refer to your notes on how to solve this. I will put this on the next slide in a picture.

Answer: .400 mole of sucrose C₁₂H₂₂O₁₁, contains _______ MOLES of C

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Question: Solving

Answer: This is how to solve question 1 and 2

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Question: 39.10 g/mole

All you have to do is look at the periodic table

Answer: The Molar mass of potassium is __________.

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Question: 95.21 g/mole

Looking at the periodic table,

Mg =24.31 and Cl = 35.45

There are 2 Cl so add 24.31+2(35.45) = 95.21

Answer: Calculate the molar mass of magnesium chloride, MgCl₂.

____________.

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Question: 74.10 g/mole

Looking at the periodic table,

Ca =40.08 , O = 16.00 and H =1.01

There are 2 O’s and 2 H’s

so 40.08 + 2(16.00)+ 2(1.01) = 74.10

Answer: The molar mass of calcium hydroxide, Ca(OH)₂, is

___________.

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Question: 0.694 g

Looking at the periodic table you see that Li has a Molar massif 6.94

So.. remember this formula as it will not be provided like the others.

m= M x n, m=mass, M = Molar mass, n = moles.

so multiply 6.94 by 0.100 and you get the answer.

Answer: 0.100 mole of lithium has a mass of _________.

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Question: 0.0861 moles

Looking at the periodic table you must find the Molar mass of K₂SO₄ as done in previous problems.

2(39.10)+32.07+4(16.00) = 174.27

Remember this formula:

n= m/M, n=moles, m=mass, M= Molar mass

so n= 15.0g/174.27 = 0.0861 moles

Answer: How many moles of K₂SO₄ are in 15.0 g of K₂SO₄?

___________.

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Question: 640. g

Looking at the periodic table you must find the Molar mass of C₆H₁₂O₆.

6(12.01)+12(1.01)+6(16.00) = 180.18

Remember the formula m=M x n

So m= 180.18 x 3.55g = 640. g

Answer: How many grams of glucose (C₆H₁₂O₆) are in 3.55 moles of glucose? ____________.

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Question: D.

Answer: Which of the following gives the balanced equation for this reaction?

K₃PO₄ + Ca(NO₃)₂ → Ca₃(PO₄)₂ + KNO₃

  1. KPO₄+CaNO₃+KNO₃

  2. B. K₃PO₄+Ca(NO₃)₂ → Ca₃(PO₄)₂+3KNO₃

  3. C. 2K₃PO₄+3Ca(NO₃)₂ → Ca(PO₄)₂ +6KNO₃

  4. D. 2K₃PO₄+ 3Ca(NO₃)₂ → Ca₃(PO₄)₂ +6KNO₃

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  6. Question: B.

  7. Answer: Which of the following gives the correct coefficients for the reaction below?

N₂H₄ + H₂O₂ → N₂+ H₂O

  1. 1,1,1,1

  2. B. 1,2,1,4

  3. C. 2,4,2,8

  4. D. 1,4,1,4

  5. E. 2,4,2,4

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  7. Question: C. 3

  8. Answer: What is the coefficient of hydrogen, H₂, when the following equation is balanced?

Al + H₂SO₄ → Al₂(SO₄)₃ + __H₂

  1. 1

  2. B. 2

  3. C. 3

  4. D. 4

  5. E. 5

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  7. Question: Single replacement

Because the iron kicked the hydrogen out.

Answer: What is the classification for this unbalanced reaction?

Fe + HCl → FeCl₃ + H₂

  1. combustion

  2. B. combination

  3. C. decomposition

  4. D. double replacement

  5. E. Single replacement

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  7. Question: C. combustion

Because the product contains water and carbon dioxide

Answer: The following reaction is an example of a ___________ reaction.

C₅H₈ + 7O₂ → 5CO₂ + 4H₂O

  1. combination

  2. B. single replacement

  3. C. combustion

  4. D. decomposition

  5. E. Double replacement

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  7. Question: D. decomposition

Because 1 product was broken down into 2

Answer: The following reaction takes place when an electric current is passed through water. It is an example of a _________ reaction.

2H₂O →(electricity)→ 2H₂ + O₂

  1. Combination

  2. B. Single replacement

  3. C. Combustion

  4. D. Decomposition

  5. E. Double replacement

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  7. Question: Pressure

  8. Answer: The force of gas particles against the walls of a container is called ___________.

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  10. Question: 760 mmHg

  11. Answer: The unit of 1 atmosphere (atm) used to describe the pressure of a gas is equal to _________.

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  13. Question: 1140 mmHg

Utilizing the formulas given you know that all this question is asking is that you convert the 1.50 atm to mmHg.

so 1.50 x 760 = 1140 mmHg

Answer: A 5.00-L tank contains helium gas at 1.50 atm. What is the pressure of the gas in mmHg? _____________.

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Question: 0.30 atm

Looking at your formula sheet you realize that this question contains 2 volume measurements and a pressure measurement; seeing the units are pressure and volume you should automatically think Boyle’s Law.

P₁V₁ = P₂V₂

realize that P₁ = 1.2 atm, V₁ = 1.0L, and V₂ = 4.0L

Now write it out in the formula substituting the values.

1.2 atm ( 1.0L) = (P₂) 4.0 L

to solve for this you must isolate the P₂ value on the right side of the equation by dividing

Answer: The final volume of a gas with a pressure of 1.2 atm. increases from 1.0 L to 4.0 L. What is the final pressure of the gas, assuming constant temperature? ___________.

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Question: 4.6 L

Realize that you are again using Boyle’s Law.

P₁V₁ = P₂V₂

write it out in the formula substituting values.

Convert atm to mmHg

keep in mind that you can also write it this way.

5.0 L (1.50 atm) = V₂(1240 mmHg)

you solve this one a bit differently.

Answer: The pressure of 5.0 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant temperature?

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Question: 4.6 L

P and V = Boyle’s Law

P₁V₁=P₂V₂

(1.5 atm)(5.0 L) = 1240 mmHg (V₂)

FIRST: convert atm to mmHg by multiplying 1.5 by 760 = 1140 mmHg

Then isolate the V₂ by dividing the 1240 mmHg into both sides.

Multiply the top then divide that answer by the bottom to get your answer in Liters.

Answer: The pressure of 5.0 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant temperature? __________.

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Question: 1170 mL

This problems units contain volume and temperature, so you use Charles’s law.

V₁/T₁ = V₂/T₂

500 ml / 150K = V₂/350K

Realize that you always want K in your problem, if you have C convert to K by adding 273.

The way to solving this problem would be to simply multiply the 500 ml by the 350K by cross multiplication then dividing that product by 150 K to get your answer in mL.

Answer: The temperature of a 500. mL sample of gas increases from 150 K to 350 K What is the final volume of the sample of gas, if the pressure in the container is kept constant? ______________ mL.

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Question: 2.1 atm

This problem is solved by using Lussac’s Law.

P₁/T₁ = P₂/T₂

1.5atm / 320K = P₂/450K

same as the last problem, just multiply P₁ by T₂ then divide that product by T₁ to get your answer in atmospheres.

Answer: A gas contained in a steel tank has a pressure of 1.5 atm at a temperature of 320 K. What will be the gas pressure when the temperature changes to 450 K? _________ atm.

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Question: 5.63 atm

This problem is solved by using Gay-Lussac’s Law.

P₁/T₁ = P₂/T₂

5.00 atm/5.0°C = P₂/40°C

FIRST CONVERT °C to KELVINS by adding 273.

5.00 atm/278K = P₂/313K

Then solve as before by multiplying P₁ by T₂ then divide the product by T₁ to get your answer.

Answer: A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0°C During the summer, the temperature in the storage area reached 40.0°C. What was the pressure in the gas tank in the summer? ________ atm.

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Question: 1270 torr.

This problem is solved by using Boyle’s Law

P₁V₁ = P₂V₂

760 torr(5.00L) = P₂(3.00L)

Isolate P₂ by dividing both sides by 3.00

Multiply the top then divide the bottom by its product.

Answer: At a constant temperature, a sample of helium at 760 torr in a closed container was compressed from 5.00 L to 3.00 L. What was the new pressure exerted by the helium on its container? _________ torr.

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Question: 5.17 L

This problem is solved by using Charles’s law

V₁/T₁ = V₂/T₂

5.00L / 25°C = V₂/35°C

FIRST convert Celsius to kelvins by adding 273.

5.00L / 298K = V₂/ 308K

Then multiply V₁ by T₂ and divide its product by T₁.

Answer: A gas sample in a closed, expandable container of initial volume 5.00 L was allowed to warm from 25°C to 35°C what was its new volume? _________ L.

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Question: no directly

Answer: Is Pressure and temperature Inversely proportionate at a fixed volume?

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