Chemistry, 15.11.2020 01:00 george6871

A chemist heats 34.10 g of nickel to 77.17 °C, then places the metal sample in the cup of water shown in the interactive. Calculate the final temperature of the water. The specific heat of nickel is 0.444 J/g °C and the specific heat of water is
4.184 J/g °C.

Answers: 1

Show answers

Answers

Answer from: Trucofer8159

$T_{eq}=27.97\°C$

Explanation:

Hello.

In this case, for equilibrium temperature problems, it is said that the heat balance allows us to notice how the hot substance heats up the cold substance until they reach the equilibrium temperature which is a temperature that remains constant upon time. Thus, since here the hot substance is the nickel and water gains that released heat by the nickel we can write:

$Q_{Zn}=-Q_{water}$

Which can be written in terms of temperatures, masses and specific heats:

$m_{Zn}C_{Zn}(T_{eq}-T_{Zn})=-m_{water}C_{water}(T_{eq}-T_{water})$

Thus, solving for the equilibrium temperature we write:

$T_{eq}=\frac{m_{Zn}C_{Zn}T_{Zn}+m_{water}C_{water}T_{water}}{m_{Zn}C_{Zn}+m_{water}C_{water}}$

Now, plugging in the known data, considering the mass of water 64.00-4.00=60.00 g and its initial temperature, 25.00°C, we obtain:

$T_{eq}=\frac{34.10g*0.444\frac{J}{g\°C}* 77.17\°C+60.0g*4.184\frac{J}{g\°C}*25.00\°C}{34.10g*0.444\frac{J}{g\°C}+60.0g*4.184\frac{J}{g\°C}}\\\\T_{eq}=27.97\°C$

Best regards!

Answer from: Quest

Ithink it’s a forms a perfect sphere or a smooth flattened oval

Answer from: Quest

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( $n_2$ ) .

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.