Answer:
Genetic factors contributes to the formation of the dead zones is by their ability to expand their abilities that enables them to spread and contribute to the formation of the dead zones. These dead zones are made when the oxygen are low where it is necessarily important for the aquatic life, if the oxygen needed is depleted or too low, instead of supporting aquatic life, dead zones are created and factors contribute to these occurrences with their ability to expand.
Explanation:
Bases
A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
Explanation:
Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.
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Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
