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Bezzdna [24]
3 years ago
15

Calculate the mass (in g) of 2.1 x 1024 atoms of W.

Chemistry
1 answer:
Monica [59]3 years ago
4 0
Atomic mass W = 183.84 u.m.a

183.84 g ----------- 6.02x10²⁴ atoms
?? g ---------------- 2.1x10²⁴ atoms

2.1x10²⁴ x 183.84 / 6.02x10²⁴ =

3.860x10²⁶ / 6.02x10²⁴ = 641.30 g

hope this helps!
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What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
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It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

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