A chemist boils off the water from 627 mL of a solution with a pH of 10.2. The only dissolved compound in the solution is NaOH. How much solid remains

4.03 x 10∧-3

Explanation:

The pH of the solution is 10.2, and we know:

pH + pOH = 14

pOH = 14 − pH

pOH = 14 − 10.2 = 3.8  

pOH = −log[OH−]

3.8 = −log[OH−]

−3.8 = log[OH−]

[OH−] = 10∧−3.8 = 1.58 × 10∧−4

since NaOH(s) → Na∧+(aq) + OH∧−(aq), [NaOH] = 1.58×10−4.

627mL × 1L/1000mL × 1.58×10−4molesNaOH/1L × 40g NaOH/1mole NaOH

=4.03 × 10∧−3g NaOH

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

the temperature of the gas, in k, not celsius.