To solve this problem we will use the following equation:
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
<span>w = mass CaCl2/(mass of water + mass of CaCl2)
</span>
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.
(a) 
It dissociates to give two particles.
(b) 
Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.
(c) 
Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.
(d) Surcose being a cobvalent compound doe not dissociate into ions. Therefore, there will be only 1 particle is present.
(e) 
Total number of particles it give upon dissociation are 1 + 1 = 2. Hence, it gives 2 particles.
Answer: The concentration of excess ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of 
give = 1 mole of 
Thus 0.019 moles of give = 0.019 mole of
2. moles of 
According to stoichiometry:
1 mole of 
gives = 2 moles of 
Thus 0.012 moles of give = 
moles of
As 1 mole of neutralize 1 mole of
0.019 mole of will neutralize 0.019 mole of
Thus (0.024-0.019)= 0.005 moles of will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of in solution is 0.017 M.
Answer:
A. There is more dissolved oxygen in colder waters than in warm water.
D. If ocean temperature rise, then the risk to the fish population increases.
Explanation:
Conclusion that can be drawn from the two facts stated above:
*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.
*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.
*Fishes, therefore, would thrive best in colder waters than warmer waters.
The following are scenarios that can be explained by the facts given and conclusions arrived:
A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)
D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).
