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Valentin [98]
3 years ago
15

Calculate the average atomic mass for element X

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

39.02amu

Explanation:

According to this question, there are four (4) isotopes for element X with the following relative abundance:

Isotope 1 = 9.67%, mass no: 38

Isotope 2 = 78.68%, mass no: 39

Isotope 3 = 11.34%, mass no: 40

Isotope 4 = 0.31%, mass no: 41

To find the average atomic mass of element X, we multiply each isotopes' relative abundance by its mass no and find the sum as follows:

We convert each percentage to decimal abundance:

Isotope 1 = 9.67% = 0.0967

Isotope 2 = 78.68% = 0.7868

Isotope 3 = 11.34% = 0.1134

Isotope 4 = 0.31% = 0.0031

(0.0967 × 38) + (0.7868 × 39) + (0.1134 × 40) + (0.0031 × 41)

3.6746 + 30.6852 + 4.536 + 0.1271

Average atomic mass = 39.02amu

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Explanation:

(a). The given data is as follows.

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Volume of 4 glass of water = 220 \times 4 = 880 ml

We known that density of water is 1 g/ml.  Therefore, calculate the mass of water as follows.

       Mass of water = 880 ml \times 1 g/ml

                               = 880 gm

                               = 0.88 Kg               (as 1 kg = 1000 g)

The relation between heat energy, mass and temperature change is as follows.

            Q = mC \Delta T

\Delta T = (37 - 3.2)^{o}C = 33.8^{o}C

Putting the given values into the above formula as follows.

         Q = mC \Delta T

             = 0.88 \times 4.186 J/g^{o}C \times 33.8^{o}C

             = 124.5 kJ

Hence, the body have to supply 124.5 kJ to raise the temperature of the water to 37 degree C.

(b).      As we know that the heat of fusion of ice is 333 J/g.

So, energy required for 8.4 \times 10^{2} g or 840 g is as follows.

          333 \times 840 = 279.72 kJ

Heat capacity of water= 4.184 J/g^{o}C

Now, heat energy will be as follows.

           Q = 4.184 \times 840 g \times 37^{o}C

               = 130.03 kJ

Therefore, total heat required = (279.72 + 130.03) kJ

                                                  = 409.75 kJ

Hence, for the given situation your body should lose 409.75 kJ  of heat.

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