Answer:
double replacement occurs
Answer:
C. It decreases by a factor of 4
Explanation:
F1 = kq1*q2/r²
F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4
The answer you are looking for is True
Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.
Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.
Thus;
Ca(CH3COO)2(aq) ----> Ca^2+(aq) + 2CH3COO^-(aq)
It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
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Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
