Flashcard maker : Amari Finch
One drop of water weighs 0.040 g. Calculate how many molecules are there in one drop, taking the molar mass of water as exactly 18 g þmol^-1.
A 1.3 × 10^21
B 2.4 × 10^22
C 3.3 × 10^22
D 3.9 × 10^22
A.
0.04g x 18 = 0.0022 moles
0.0022 x 6.02×10^23 = 1.337×10^21
Calculate to 3 sig figs the mass of 1 ? 4 mole of Li2O.
7.468g
6.94+6.94+15.99 = 29.87 g/mol
29.87/4 = 7.468
Calculate to 3 sig figs the mass of 2.60 × 10^22 molecules of dinitrogen
monoxide.
1.90g
6.02 / .260 = 23.153
(14+14+16) / 23.153 = 1.90g
Calculate how many moles are in 4000g of Nickel Sulfate
25.848 moles
4000g / (58.6934 +32.06 + (16×4)) = 25.848
3.01 × 10^25 molecules of a gas has a mass of 6.40 kg. Determine its molar mass.
128 g/mol
6.4×1000 = 6400g
6.02/301 = 0.02
6400 x 0.02 = 128 g/mol
Of the following, the only empirical formula is
A C12H22O11
B c6h12o6
C C6H6
D C2H4
A.
Determine the percentage by mass of silver in silver sulfide (Ag2S).
A 33.3%
B 66.7%
C 77.1%
D 87.1%
D.
107.86 x 2 + 32.06 = 247.78 g/mol
107.86 x 2 = 215.72
215.72/247.78 = 0.87106 = 87.1%
2.40 g of element Z combines exactly with 1.60 g of oxygen to form a compound with the formula ZO2. Determine the relative atomic mass of Z.
A 24.0
B 32.0
C 48.0
D 64.0
C.
1.6 g / 16 = 0.1
0.1 / 2 = 0.05
0.05 x 2.4 = 48g/mol
A certain compound has a molar mass of about 56 g mol-1. All the following are possible empirical formulas for this compound except:
A CH2
B CH2O
C c3h4o
D CH2N
B.
CH2O is the only one with a molar mass that isn’t a factor of 56.
When magnesium is added to aqueous silver nitrate, the following reaction takes place
Mg(s) + 2 AgNO3 (aq) -> 2 Ag(s) + Mg(NO3)2 (aq)
What mass of silver is formed when 2.43 g of magnesium is added to an excess of aqueous silver nitrate.
A 107.9 g
B 21.6 g
C 10.8 g
D 5.4 g
B.
2.43 / 24.305 g/mol = 0.09998
0.09998 x 2 = 0.199958
0.199958 x 107.868 = 21.569 g
Freon-12 (used as coolant in refrigerators), is formed as follows:
3CCl4(l) +2SbF3(s) -> 3CCl2F2(g) +2SbCl3(s)
150 g tetrachloromethane is combined with 100 g antimony(III) fluoride to give Freon-12 (CCl2F2).
(a) Identify the limiting and excess reagents.
(b) Calculate how many grams of Freon-12 can
be formed.
(c) How much of the excess reagent is left over?
a) Limiting reagent is SbF3, CCl4 is in excess
b) 101.5 g
c) 20.9 g

a) CCl4: 150g/156gmol = 0.987
0.987 / 3 = 0.329
SbF3: 100g / 178.7 gmol = 0.2795
0.2797 / 2 = 0.2795
SbF3 is limiting

b) 100g x 1/178.7gmol x 3/2mol x 120.7gmol = 101.5g

c) 100g x 1/178.7gmol x 3/2mol x 153.8 = 129.09g
150g-129.09g = 20.9 g

What volume would 3.20 g of sulfur dioxide occupy at s.t.p.?
A 0.56 dm3
B 1.12 dm3
C 2.24 dm3
D 4.48 dm3
B.
3.2g / 64.06gmol = 0.04995
0.04995 x 22.4 = 1.118dm = 1.12dm
A mixture of 20 cm3 hydrogen and 40 cm3 oxygen is exploded in a strong container. After cooling to the original temperature and pressure (at which water is a liquid) what gas, if any, will remain in the container?
30cm unreacted.
In water, hydrogen and oxygen has a 2:1 ratio. Thus, 20cm of H reacts with 10cm of O, leaving 30cm behind.
To three significant figures, calculate how many methane molecules are there in 4.48 dm3 of the gas at standard temperature and pressure?
1.20×10^23
4.48/22.4 = 0.2
0.2 x 6.02×10^23 = 1.20×10^23
If 3.00 dm3 of an unknown gas at standard temperature and pressure has a mass of 6.27 g, Determine the molar mass of the gas.
46.816 gmol
3 / 22.4 = 0.13392
6.27g / 0.13392mol = 46.816 gmol
A sealed flask contains 250 cm3 of gas at 35C and atmospheric pressure. The flask is then heated to 350C. The pressure of the gas will increase by a factor of about
A 2
B 10
C 250
D 585
A.
Sodium phosphate has the formula Na3PO4. Determine the concentration of sodium ions in a 0.6 mol dm-3 solution of sodium phosphate?
A 0.2 mol dm-3
B 0.3 mol dm-3
C 0.6 mol dm-3
D 1.8 mol dm-3
D.
There are 3 Na ions in ever sodium phosphate so..
0.6 x 3 = 1.8
What volume of a 0.5 mol dm-3 solution of sodium hydroxide can be prepared from 2 g of the solid?
A 0.05 litres
B 0.1 litres
C 0.4 litres
D 0.5 litres
B.
2g / 40gmol = 0.05
0.05 mol / 0.5M = 0.1 liters
500 cm3 of 0.500 mol dm-3 NaCl is added to 500 cm3 of 1.00 mol dm-3 Na2CO3 solution. Calculate the final concentration of Na+ ions in solution.
1.25 mol dm-3
500/1000 = 0.5dm
0.5 x 0.5M = 0.25
0.5 x 1M = 0.5
0.5 x 2= 1
1+0.25 = 1.25 mol dm-3
How would you prepare 1.2 dm3 of a 0.40 mol dm-3 solution of hydrochloric acid starting from a 2.0 mol dm-3 solution?
Measure out 240 cm3 of 2.0 mol dm-3 hydrochloric acid (0.48 moles) and make this up to 1.2 dm3 of solution.
1.2 x 0.4 = 2 x ?
0.48 = 2?
? = 0.24dm = 240cm