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Paraphin [41]
4 years ago
9

The volume of water in a transfer pipette is 15.23 ml. A 6.7 mL

Chemistry
1 answer:
Alja [10]4 years ago
3 0

Given :

Volume of water in a transfer pipette is 15.23 ml.

A 6.7 ml  volume of water was then transferred out.

To Find :

The new volume.

Solution :

Initial volume , V_i=15.23\ ml.

Volume removed out , v = 6.7 ml .

Let, new volume be V_f.

Now , new volume = Initial volume - Volume removed out .

V_f=V_i-v\\\\V_f=15.23-6.7\\\\ V_f=8.5\ ml

Therefore, new volume is 8.5 ml.

Hence, this is the required solution.

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Compounds like CCl₂F₂ are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are n
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Explanation:

The given data is as follows.

        Specific heat of water = 4.18 J/g^{o}C

        Heat of fusion of water = 334 J/g

        Mass of water = 200 g

On bringing water at 0^{o}C, heat released will be as follows.

            q_{1} = m \times C \times \Delta T

                    = 200 g \times 4.18 J/g^{o}C \times (0 - 15)^{o}C

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or,                = -12.540 kJ           (as 1 kJ = 1000 J)

Now, calculate the heat releasedwhen water freezes at 0^{o}C as follows.

                q_{2} = mass \times \text{-heat of fusion}

                         = 200 g \times -334 J/g

                         = -66800 J

or,                      = -66.80 kJ

Therefore, total heat released in freezing water will be as follows.

                     q_{total} = q_{1} + q_{2}

                              = (-12.540 - 66.80) kJ

                              = -79.34 kJ

Hence,

       amount of heat released in freezing water = heat used to vaporize CCl_{2}F_{2}

Now, heat of vaporization of CCl_{2}F_{2} = 289 J/g

Total heat released in freezing water = -79.34 kJ

Heat consumed to vaporize CCl_{2}F_{2} = 79.34 kJ = -79340 J

Therefore, calculate the mass of CCl_{2}F_{2} vaporized as follows.

              Mass of CCl_{2}F_{2} vaporized = \frac{\text{heat consumed}}{\text{heat of vaporization}}

                              = \frac{79340 J}{289 J/g}

                              = 274.53 g

Thus, we can conclude that 274.53 g mass of this substance must evaporate to freeze 200 g of water initially at 15^{o}C.

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