The density of a metal is 11.4 g/cm3. How much volume (in cm3) would a sample of 30.5 g have?
1 answer:
<u>The Concept:</u>
We are given the density of a sample of the metal = 11.4 grams / cm³
and we need to find the volume occupied by a sample of 30.5 grams
For this solution, we will use dimensional analysis
from the given information, we can also say that the density of the metal is:
1 cm³ / 11.4 grams
If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal
<u>Solving for the volume:</u>
X 30.5 grams = (30.5 / 11.4) cm³
Volume of 30.5 grams of the sample = 2.68 cm³
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Answer:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce = 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g