Answer:
The answer you have selected in the screenshot is correct.
Its tendency to react with oxygen is correct.
Hope this helps.
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
Answer:
In numerical order left to right, they are arranged by the number of protons in the nucleus of a single atom of each element
Answer:
c. 2,2-dichloropentane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:
Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:
Whose name is 2,2-dichloropentane.
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