What would the expected temperature change be (in fahrenheit) ida 0.5 gran sample of water released 50.1 j of heat energy? the specific heat of liquid water is 4.184 j/g-c

23.95 °C

We are given;

We are required to calculate the change in temperature;

That is, Q = mcΔT

Rearranging the formula;

ΔT = Q ÷ mc

Therefore;

ΔT = 50.1 J ÷ (0.5 g × 4.184 J/g°C)

    = 23.95 °C

Therefore, the expected change in temperature is 23.95 °C

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.

beta decay is the answer