(a) 12 cm + 0.031 cm + 7969 cm =
2 answers:
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Answer:
C）52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
;
- calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.