A substance with a density of 2.70 g/mL occupies a volume of 21.3 mL. What is the mass of the sample? Report your answer in unit
1 answer:
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Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles =
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = = 0.03mole
Number of moles of H₂O = = 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2
Explanation:
When pH of the solution is 11.
..(1)
At pH = 11, the concentration of ions is
.
When the pH of the solution is 6.
..(2)
At pH = 6, the concentration of ions is
.
On dividing (1) by (2).
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is .
Difference between the ions at both pH:
This means that Hydrogen ions in a solution at pH = 7 has ions fewer than in a solution at a pH = 6