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IRISSAK [1]
3 years ago
10

A substance absorbed 895 J and caused its temperature to increase by 10 ºC. If the substance has a mass of 27.9 g, what is the s

ubstance's specific heat in J/(gºC) ?
1.) 3.21

2.) 4.62

3.) 1,406.25

4.) 249,705
Chemistry
2 answers:
Irina18 [472]3 years ago
5 0

Answer:

i dont get the equation sorry wish i can help

vova2212 [387]3 years ago
4 0
4.62 is the correct answer
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Nitrogen gas in an expandable container is cooled from 45.0 C to 12.3 C with the pressure held constant at 2.58 ∗ 105 Pa. The to
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Answer:

The number of moles of the gas is: -27.14 mole

the charge in internal energy of the gas is -1.84 × 10⁴ J.

The work done by the gas is 4.42 × 10⁴ J

The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

Explanation:

The expression for the number of moles of a gas at constant pressure is as follows:

\mathbf{n = \frac{Q}{Cp \Delta T}}

\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}

where ;

C_p is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K

Since heat is liberated from the gas ; then:

n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}

n = -27.14 mole

The number of moles of the gas is: -27.14 mole

b) The expression to be used in order to determine the change internal energy is:

dU = nCv \Delta T

where ;

n= 27.14 mole

Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K

ΔT = (12.3-45)

So;

dU = (27.14)(20.76)(12.3-45)

dU = 563.426(-32.7)

dU = -18424.04328

dU = -1.84 × 10⁴ J

Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.

c)  The workdone by the gas can be calculated as;

W = Q - ΔU

W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )

W = 2.58  × 10⁴ J + 1.84  × 10⁴ J

W = 4.42 × 10⁴ J

The work done by the gas is 4.42 × 10⁴ J

d) The expression to calculated the work done is given as:

W = pdV

since the volume is given as constant ; then dV = 0

so;

W = p(0)

W = 0

Replacing 0 for W in the equation W = Q - Δ U

0 = Q - ΔU

-Q = - ΔU

Q = ΔU

Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

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Plug in 33 * 0.5 M = 45 *x, where x is the concentration of h3po3, solve for x
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