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andrezito [222]

2 years ago

12

How many particles are in 100.0 grams of gold

1 answer:

Tcecarenko [31]2 years ago

3 0

First, we need to convert the grams of gold into mole using the molar mass (molecular weight) of gold, and then into particles using avogadro's number (1 mol= 6.02 x 10^23)

molar mass of gold= 197 grams/ mole

100.0 grams (1 mole/ 197 grams) x (6.02 x 10^23 particles/ 1 mole) = 3.06 x 10^23

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The bomb calorimeter in Exercise 102 is filled with 987g water. The initial temperature of the calorimeter contents is 23.32. A

Sholpan [36]

Answer:

25.907°C

Explanation:

In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C

The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.

Thus:

$-q_{combust} = q_{water} + q_{calori}$

$q_{combust}$ = heat of combustion of benzoic acid

$q_{water}$ = heat energy released to water

$q_{calori}$ = heat energy released to the calorimeter

Therefore,

$-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})$

1.056*26.42 = [0.987*4.18 + 6.66]( $T_{f}$ - 23.32)

27.8995 = [4.12566+6.660]( $T_{f}$ - 23.32)

( $T_{f}$ - 23.32) = 27.8995/10.7857 = 2.587

$T_{f}$ = 23.32 + 2.587 = 25.907°C

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