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).
Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
Answer:
B is a precipitation reaction.
Explanation:
This is because a precipitation reaction is when a solid is made from the combination of cations and anions in a solution to create a solid.
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
It is achievable for the same enzyme to catalyze reverse reactions for the reason that the direction of a reversible reaction is determined by the concentrations of reactants and products. In pulmonary circulation, the low CO2 concentration supports the making of CO2 and H2O.