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Chemistry 11 Assignment 2 Total 25 marks

[pic 1]

Please make sure you highlight your final answer. Failure to do so will result in one mark taken off.

Part 1

For the next questions, clearly show all the steps in solving the problems.  0.5 marks will be deducted for each time there is a significant figure error or appropriate units are not used!

1.  What is the molar mass (to the nearest tenth of a gram/mole) of each of the following (2 marks)

a. KCl

K: 1 x 39.1 = 39.1

Cl: 1 x 35.5 = 35.5

39.1 + 35.5 = 74.6 g/mol

b. Al2(SO4)3

Al: 2 x 27.0 = 54.0

S: 3 x 32.1 = 96.3

O: 12 x 16.0 = 192

96.3 + 54.0 + 192 = 342g/mol

2.  Convert the following mass into moles (1 mark)

   19.5 grams of Al(OH)3

Al: 1 x 27.0 = 27.0

O: 3 x 16.0 = 48.0

H: 3 x 1.0 = 3.0

27.0 + 48.0 + 3.0 = 78.0

19.5 grams Al(OH)3 x 1 mole / 78.0 grams Al(OH)3 =  0.250 mol Al(OH)3

3.  Convert the following moles into masses (1 mark)

    2.00 moles of N2

2.00 moles x 28.0 g / 1 mole =56.0g N2

4. How many molecules are present in 1.5 moles of NaCl? (1 mark)

1.5 moles NaCl x 6.02 x 10 ^23 molec NaCl/ 1 mole NaCl = 9.0 x 10^23 molec NaCl

5.  How many atoms of chlorine are in 19.0 grams of MgCl2? (1 mark)

Mg: 1 x 24.31 = 24.31

Cl: 2 x 35.45 = 70.9

24.31 + 70.9 = 95.21       95.2 g/mol[pic 2]

19.0 g MgCl2 x 1 mol MgCl2 /95.2 g Mg Cl2 x 6.02 x 10^23 molec/ 1 mol x 2 atoms/ 1 molec = 2.40 x 10 ^23 atoms

6.  What is the volume of 2.60 moles of CO2(g) at STP? (1 mark)

2.60 mol CO2 x 22.4 L CO2 / 1 mol CO2 = 58.24 L CO2         58.2 L CO2[pic 3]

7.  What is the mass of 50.0 L of O2(g) at STP?  (1 mark)

50.0 L O2 x 1 mol O2 / 22.4 L O2 x 15.99 g O2  / 1 mol O2 =35.7 g O2

8.  What is the percent composition of chlorine in PCl5?  (1 mark)

Cl: 5 x 35.45= 177.25

P: 1 x 30.97 = 30.97

177.25 + 30.97 = 208.22        

Cl: 177.25/208.22 x 100% = 85.13 %        

The percent composition of Chlorine is 85.13 %       

9.  Steven analyzed a compound and found it to be 52.7% Potassium and 47.3% Chlorine.   He determined the empirical formula to be …    (1 mark)

52.7 g K x 1 mole K / 39.1 g k = 1.35 mol K/ 1.33 ≈ 1

47.3 g Cl x 1 mole Cl/ 35.45 g Cl = 1.33 mol Cl/1.33 = 1

Empirical formula: KCl

10.  If the empirical formula of a compound is CH2 and its molar mass is 70.0, then what is its molecular formula? (1 mark)

12.0 C + 2 x 1.0 H = 14.0g/mol

70.0 g/mol ÷ 14.0g/mol = 5

5 x CH2 = C5H10

11.  If 0.30 mol NaCl goes into a solution with 4.00L water, the molarity of NaCl is (2 marks)

M = n / V  

M= 0.30 mol / 4.00L

M= 0.0750 mol/ L 

12.  If 100.0 ml water is added to 150.0 ml of a 2.00M HCl solution, the new molarity of HCl is (2 marks)


150.0 mL x 1 L/ 1000 mL = 0.1500L

100.0Ml x 1L/1000mL= 0.1000L

n = mv

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