Answer: 20) 2365 g
21) 22.39 grams.
22) 29.99 g
Explanation: 20) molarity is the no of moles of solute per unit volume.
We can calculate amount of CaCl2 required to prepare 0.1 M CaCl2 1000 ml solution.
we know that to prepare one ltr of 1 M solution of CaCl2 111 g required
Now consider x gram will require to prepare to
so that comparing above both condition
1000ml ×1M×X g=1000ml×0.1M×111g
X= 11.1 gram
X= 11.1 g of CaCl2
Hence 11.1 g of CaCl2 would be dissolved in 1.0L of a 0.100 M solution of CaCl2
21) How many moles of CaCl₂ in that solution?
;
;
.
What's the mass of that 0.20172 moles of CaCl₂?
Molar mass from a modern periodic table:
Ca- 40.078;
Cl- 35.45.
Molar mass of CaCl₂:
.
Mass of that 0.20172 moles of CaCl₂:
22) its a 3.0m solution so 1 litre of solution contains 3 moles of NaOH, 250ml of solution contains 0.25x39.9971 g/mol, so 250ml of this solution contains 0.75x39.9971=29.99g, or if you round it up 30.0g
Answer:
V= 0.147 L
Explanation:
This is simply the application of combined gas law twice, to find the unknowns.
Combined gas law states that: 
P= pressure of air
V= volume of air
n= moles of air
R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)
T= Absolute temperature in kelvin.
Now, applying the same gas law at 483K and substituting for n
V= 
V= 0.147 L
Answer:
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