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lisov135 [29]
3 years ago
12

The temperature of a 500. ml sample of gas increases from 150. k to 350. k. what is the final volume of the sample of gas, if th

e pressure and moles in the container is kept constant? 0.0095 ml 110. ml 0.0470 ml 210. ml 1170 ml
Chemistry
2 answers:
Andreas93 [3]3 years ago
7 0
<span>Answer: 1170 ml is the final volume of the sample of gas, if the pressure and moles in the container is kept constant.</span>
Talja [164]3 years ago
7 0

Answer:

The final volume of the sample of gas is  1170 L.

Explanation:

Initial volume of the gas =V_1= 500 mL= 0.5L

Initial temperature of the gas =T_1= 150 K

Final volume of the gas =V_2=?

Final temperature of the gas =T_2=350 K

Since, pressure and number of moles remains constant we can apply Charles law:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

\frac{0.5 L\times 350 K}{150 K}=V_2

V_2=1.16666 L=1166.66 mL\approx 1170 L

The final volume of the sample of gas is 1170 L.

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If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coeff
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Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

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6 0
4 years ago
179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g
taurus [48]

Answer:

the specific heat of the unknown compound is c_u=0.412J/g \cdot C

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Generally the change in temperature of water is evaluated as

                \Delta T = T_2 -T_1

Substituting 16.1°C for T_1 and 27.4°C for T_2

                \Delta T = 27.4 - 16.1

                       =11.3^oC

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                  \Delta T_u = T_3 -T_2

Substituting 27.4°C for T_2 and 94.3°C for T_3

                                    \Delta T = 94.3 - 27.4

                                           =66.9^oC

Since there is an increase in temperature then heat is gained by water and this can be evaluated as

               H_w = mc_w \Delta T

Substituting 179.1 g  for m , 4.18 J/g.C for c_w(specific heat of water)

             H_w = 4.18 * 179.1 * 11.3

                   = 8459.6J

Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

                    H_u = m_uc_u \Delta T_u

By conservation of energy law

       Heat lost  = Heat gained  

Substituting 306.9 g  for m_u , 8459.6J for H_u

           8459.6 = 306.9 * c_u * 66.9

  Therefore     c_u = \frac{8459.6}{308.9 *66.9}

                           =0.412J/g \cdot C

                   

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Explanation:

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