What type of reaction occurs when potassium iodide is combined with lead (II) nitrate?
1 answer:
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The mass of the air in the room is 50 kg.
V = lwh = 15.5 ft × 12.5 ft × 8 ft = 1550 ft³
1 ft = 12 in ×
= 30.48 cm
V = 1550 ft³ ×
= 4.39 × 10⁷ cm ³ = 4.39 × 10⁷mL = 4.39 × 10⁴ L
Mass = 4.39 × 10⁴ L ×
= 5.22 × 10⁴ g = 52.2 kg
Note: The answer can have only 1 significant figure, because that is all you gave for the height of the room.
To the correct number of significant figures, the mass of the air in the room is 50 kg.
V = lwh = 15.5 ft × 12.5 ft × 8 ft = 1550 ft³
1 ft = 12 in ×
V = 1550 ft³ ×
Mass = 4.39 × 10⁴ L ×
Note: The answer can have only 1 significant figure, because that is all you gave for the height of the room.
To the correct number of significant figures, the mass of the air in the room is 50 kg.
Answer:
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula