A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the substance was 125.0 °C. Determine the specific heat of this substance.

0.296j/g⁰c

Explanation:

we have the following information from this question before us.

mass iv substance = 50grams

we have initial temperature ti = 20.2⁰c

final temperature = 125⁰c

the energy that was provided = 155kj

we proceed with this formula

energy = mcΔT

1.55x10³ = 50 x c x (125-20.2)

1.55x10³ = c x 50gm x 104.8k

we divide through to get c

c = 1.55x10³/50g x 104.8

c = 0.296J/g⁰c

that is the specific heat of this substance.

thank you!

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.

14,16 18 reacts with group