Apure gold ring with a volume of 1.79 cm^3 is initially at 17.4 ∘c. when it is put on, it warms to 28.5 ∘c. how much heat did the ring absorb? (density of gold =19.3 g/cm^3)

Hi Adriana!. You should need to find out which is the specific heat for gold. As this value is 0.129 J/g °C.  The ring has absorbed 49.5 J of heat.

Explanation:

First of all, you have to use the density formula to find out your mass.

d = mass/volume

19.3 g/cm3 = mass / 1.79 cm3

19.3 g/cm3 x 1.79 cm3 = mass

34.55 g = mass

Now, that we have the gold mass, let's go to the specific heat formula

Heat = mass . Specific heat . ΔT (where ΔT means the difference between temperatures.- Tfinal - Tinitial) So..

Heat = mass . Specific heat . ΔT

Heat = 34.55 g x 0.129 J/g °C x (28.5°C - 17.4°C)

Heat = 34.55 g x 0.129 J/g °C x (11.1°C)

Heat = 49.5 J

the answer is definitely c.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-