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Calvin told Marie that they could continue to add solute until the reached 40 grams because the solution was still unsaturated.
Unsaturated solutions are those in which the solvent (in this case water) can still dissolve more solute (in this case KNO₃) at the given pressure and temperature. This can be seen visually when adding more solute doesn't result in the presence of grains of solids that settle in the bottom of the flask. That happens because the rate of dissolving is higher than the rate of crystallization.
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Answer
D 160g
Explanation:
<u>Write the equation:</u>
Combustion reactions use oxygen and release water and heat, so
CH₃OH(g) + O₂(g) → CO₂(g) + H₂O(g)
Balance that:
2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
<u>Find moles of carbon dioxide:</u>
We need to know the number of moles of CO₂. This rxn is at STP, so at STP one mole of gas = 22.4 liters.
112 L * 1 mol/22.4 L = <em>5 mol CO₂</em>
<u>Find moles of methanol:</u>
Based on the chemical equation, for every 2 mol methanol, there are 2 mol carbon dioxide. So for every 5 mol carbon dioxide, there are 5 mol methanol!
5 mol CO₂ = 5 mol CH₃OH
Molar mass of methanol: 12.01 + 3*1.008 + 16.00 + 1.008 = <em>32.04 g/mol</em>
Moles of methanol: 5 mol * 32.04 g/mol = 160.2 g methanol
≈ 160 mol methanol
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Answer:
(1) separate the substances
(2) chemically combine the substances
(3) determine the freezing point of the mixture
(4) predict the electrical conductivity of the mixture
Explanation:
Answer:
= 82%
Explanation:
Percentage purity is calculated by the formula;
% purity = (mass of pure chemical/total mass of sample) × 100
In this case;
1 mole of Ca(NO3)2 = 164 g
but; 164 g of Ca(NO3)2 = 40 g Ca
Therefore; mass of Ca(NO3)2 = 164 /40
= 4.1 g
Thus;
% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100
= (4.1 g/ 5 g) × 100
= 82%
