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3 years ago

9

A pure 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 t

he ring absorb? (density of gold =19.3 g/cm^3)

1 answer:

yKpoI14uk [10]3 years ago

6 0

Answer:

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

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To find the number of moles, given the mass of the solute, we use the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n = number\:of\:moles\:(mol)}$

$\mathrm{m = mass\:of\:solute\:(g)}$

$\mathrm{M = molar\:mass\:of\:solute\:( \dfrac{ g }{ mol } )}$

Label the variables with the numbers in the problem:

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M =\:?\:Calculate\:the\:molar\:mass }$

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

$\mathrm{M =molar\:mass }$

$\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}$

$\mathrm{Na =sodium=22.99\:g }$

$\mathrm{S =sulfur=32.06\:g }$

$\mathrm{O =oxygen=16.00\:g }$

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$M = 2 \left( 22.99 \right) +1 \left( 32.06 \right) +4 \left( 16.00 \right)$

$\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}$

$\mathrm{M = 45.98+32.06+4\:(16)}$

$\mathrm{Multiply\:4\:by\:16\:to\:get\:64}$

$\mathrm{M = 45.98+32.06+64}$

$\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}$

$\mathrm{M = 78.04+64}$

$\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}$

$\mathrm{M = 142.04}$

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M = 142.04\:g/mol}$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$\mathrm{n = \dfrac{ 30 }{ 142.04 }}$

$\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}$

$\mathrm{n = 0.21120811}$

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

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