Answer:
Percent by mass of water is 56%
Explanation:
First of all calculate the mass of hydrated compound as,
Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g
Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g
Mass of Oxygen = O × 14 = 16 × 14 = 224 g
Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g
Mass of Na₂S0₄.10H₂O = 322.24 g
Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...
Mass of water = 10 × 18.02
Mass of water = 180.2 g
Now, we will apply following formula to find percent of water in hydrated compound,
%H₂O = Mass of H₂O / Mass of Hydrated Compound × 100
Putting values,
%H₂O = 180.2 g / 322.24 g × 100
%H₂O = 55.92 % ≈ 56%
Answer:The molar mass of atoms of an element is given by the standard relative atomic mass of the element multiplied by the molar mass constant, 1 × 10−3 kg/mol = 1 g/mol.
Explanation:
Answer: when concentrations of acid and base are same, pH = pKa
PH = 12.38 pOH = 1.62
Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00
The collision of the molecules between the hydrogen molecule or H2, and an iodine molecule or I2, provided there would be a sufficient energy is that the system would eventually undergo a chemical change wherein a new chemical compound would be formed from these two molecules.
