Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 10
Unit 2 Revision
Name: ………………………………………. Date: ……………………………………………………
Write the scientific term:
1. Expresses the chemical symbols and formulas of the reactants and products.
The chemical equation
2. The smallest part of a substance that can be found in a single state. The properties of
the substance are evident in it.
The molecule
3. The smallest building unit of the substance that participated in chemical reactions.
The atom
4. The mass of one atom.
The atomic mass
5. The mass of one molecule in grams.
The mole
6. The sum of atomic mass of the atoms forming that molecule.
The molecular mass
7. The amount of one of the reactants when it is less than its molar ratio in the balanced
chemical equation.
The limiting reactant
8. The number of atoms, molecules or ions found in one mole of the substance and
equals 6.02 ✕ 1023 (atom, molecule or ion).
Avogadro’s number
9. The amount of matter that contains Avogadro’s number of molecules.
The mole
10.The volume of gas involved in the reaction and produced from it both are in a fixed
ratios.
Gay – Lussac’s Law
11.The equal volumes of different gases contain the same number of molecules under
the same standard temperature and pressure.
Avogadro’s Law
12.The number of units from the particle for each 100 units from the overall.
Weight percentage
13.A formula expressing the simplest ratio of true numbers between the atoms of
elements which formed the compound.
The empirical formula
14.A symbolic formula of the molecule of the element, or molecule or the formula unit.
It expresses the actual type and number of atoms or ions that form this molecule or
unit.
Chemical formula
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Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 10
Unit 2 Revision
15.The amount of a substance which we practically obtain from a reaction
The practical yield
16.The amount of the calculated substance based on the reaction's equation.
The theoretical yield
Give reason for the following:
1. The volume occupied by 26 g of acetylene C2H2 at (STP) is equal to the volume
occupied by 2 g of hydrogen at the same conditions.
Because 26 grams of acetylene and 2 grams of hydrogen represent 1 mole and one mole of
any substance have the same volume
2. The difference in the molar mass of phosphorus by the difference in its physical
state.
Because when phosphorus is in solid state the molecule consists of 1 atom only, while
when it is in vapor state the molecule consists of 4 atoms
3. One liter of oxygen gas contains the same number of molecules that one liter of
chlorine gas contains at STP.
Because according to Avogadro’s Law equal volumes of different gases contains the same
number of molecules.
4. The number of molecules of 9 g of water (H2O) is equal to the number of molecules
of 39 g aromatic benzene C6H6.
Because 9 grams of water and 39 grams of aromatic benzene = 0.5 mole and according to
Avogadro’s number one mole of any substance contains the same number of molecules
5. The chemical equation should be balanced.
Because according to the law of conservation the number of atoms in reactants must equal
the number of atoms in the products
6. When you calculate the gas volume in terms of its molar mass, it should be placed in
the standard conditions of pressure and temperature.
Because equal volumes contain the same number of molecules only under standard
temperature and pressure
7. The practical yield is always less than the theoretical yield of the equation.
Some of products evaporate – some of the products clink on the walls of the test tubes or
beakers – some products may not be pure enough
8. The molar mass of the sulfur in the solid state differs from its molar mass in the
gaseous state.
Because in solid state the molecule consists of 1 atom only while in gaseous state the
molecule consists of 8 atoms
Choose the correct answer:
1. Number of water moles found in 36 g of it are (0.5 – 1 – 2 – 2.5) moles.
2. Number of carbon dioxide molecules found in 128 g of it equals (2 –
6.022 × 1023 – 3.011× 1023 – 18.044 × 1023) molecules.
3. Number of produced sodium ions from dissolving 40 g of NaOH in
water equals (2 – 6.022 × 1023 – 3.011× 1023 – 18 × 1023) ions.
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Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 10
Unit 2 Revision
4. The volume of 4 g of hydrogen molecules H2 in standard conditions (STP) equals (2
– 22.4 – 44.8 – 89.6) liter.
5. The volumes of producing gasses from the reaction are directly proportional with the
participating gas volumes in the reaction (Avogadro’s Law - Avogadro’s number Jay-Lussac Law - Law of Mass Reservation).
6. The number of the empirical formula units of the compound C2H2O4 is (1 – 2 – 3 –
4).
7. The mass of CaO resulted from dissociating 50 g of calcium carbonate CaCO3
thermally is (28 – 82 – 96 – 14) grams.
8. The number of hydrogen moles needed to produce 11.2 L of water vapor at (STP) is
(22.4 – 44.8 – 0.5 – 68.2) liter.
9. If the empirical formula of a compound is CH2 and its molecular mass is 56 , then
the molecular formula of such a compound is (C3H6 - C2H4 – C4H8 – C5H10).
10. The masses of atomic particles are estimated by the unit of atomic masses (a.m.u)
which equals (one atomic mass unit equals ????? in grams) (6.022 × 10 23 – 1.66 ×
1023 – 1.66 × 10-23 – 1.66 × 10-24) grams. Calculate in Kg.
11.The unit used in the international system SI to express the amount of a substance is
(gram – kilogram – mole – unit of atomic masses).
12.The number of grams of 44.8 L of ammonia gas NH3 at (STP) equals (2 – 17 – 34 –
0.5) grams.
13.If an amount of sodium contains 3.01 ✕ 1023 atoms, then the mass of this amount
equals (11.5 – 23 – 46 – 0.5) grams.
14.A chemical equation is to be balanced to satisfy (Avogadro - Reservation of energy Reservation of mass - Gay & Lussac).
15.Half a mole of carbon dioxide CO2 is (44 – 22 – 88 – 66).
16.The empirical formula of CH2O expresses the molecular formula of (HCHO CH3COOH - C6H12O6).
17.When 64 g of oxygen reacts with abundance of hydrogen, then the volume of the
water vapor resulted in STP is (22.4 - 44.8 - 11.2 - 89.6).
Solve the following problems:
First: Calculate the molecular formula for a compound contains carbon with a ratio 85.7 %
and hydrogen with a ratio 14.3 % and its molecular mass is 42.
C
H
85.7
14.3
12
1
7.14
14.3
1
2 the empirical formula CH2
The number of empirical units = molecular mass / empirical mass = 42/14 = 3
The chemical formula C3H6
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Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 10
Unit 2 Revision
Second:130 g of silver chloride precipitated when a mole of sodium chloride dissolved in
water reacted with silver nitrate solution. Calculate the following:
.1
The percentage of the actual product
The percentage of the actual yield = (the actual yield \ theoretical yield) * 100
= (130/143.5) *100 = 90.59%
.2
The number of sodium ions resulted from this reaction.
AgNO3 + NaCl → AgCl + NaNO3
1 mole of sodium ions is produced then the number of sodium ions = Avogadro’s number
= 6.022 × 1023
Third: Calculate the number of moles of 144 g of carbon.
1 mole of carbon
12 grams
How many moles 144 grams
How many moles = 144/12 = 12 moles
Fourth: Calculate the mass of 2.4 mole of a limestone CaCO3
1 mole of CaCO3
100 grams
2.4 mole
how many grams
How many grams = 2.4 × 100 = 240 gram
Sixth: Calculate the volume of 56 g of nitrogen at (STP).
The mass of 1 mole of nitrogen
14 gram
How many moles
56 grams
How many moles = 56/14 = 4 mole
The volume of gas = the number of moles × 22.4
= 4 × 22.4 = 89.6 Liter
Seventh: Calculate the volume of hydrogen and the number of sodium ions resulted from
the reaction of 23 g of sodium and an excess amount of water in the standard conditions
according to the following equation:
2Na + 2H2O → 2NaOH + H2
4 moles of hydrogen atoms
The volume of gas = the number of moles ×22 .4
= 4 × 22.4 = 89.6 liter
Eighth: Calculate the volume of one mole of phosphorus in the gaseous state at (STP), and
then calculate the number of the atoms in this volume.
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Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 10
Unit 2 Revision
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Ninth: 39.4 g of barium sulfate BaSO4 precipitated when 40 g of barium chloride solution
BaCl2 reacted with an abundance of potassium sulfate. Calculate the percentage of the
practical yield of barium sulphate.
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Tenth: Calculate the ratio of iron present in the raw siderite FeCO3.
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Port Said International Schools
Better Education for Future Generations
Science Department
http://pisscience.wikispaces.com/
Grade: 10
Unit 2 Revision