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2 years ago

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A piece of metal is cooled in a water calorimeter. This process increases the water temperature by 7.0°C. How much energy is los

t if there is 100. g of water in the calorimeter? (specific heat of water is 4.184 J/g°C)

1 answer:

vova2212 [387]2 years ago

3 0

Heat changes

Q = m.c.Δt

Q = 100 g x 4.184 J/g°C x 7 °C

Q = 2928.8 J

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A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

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<u>Answer:</u> The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

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Considering the reaction as:

$Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3$

1 moles of $Ba(NO_3)_2$ react with 1 mole of $H_2SO_4$

Thus,

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$2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}$

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