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lilavasa [31]
3 years ago
8

Two tiny particles carrying like charges of the same magnitude are apart. if the electric force on one of them is what is the ma

gnitude of the charge on each of these particles?(k = 1/4πε0 = 9.0 × 109 n • m2/c2)
Chemistry
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

2.4\times 10^{-8} C

Explanation:

We are given that

Distance between two tiny particles=d=1mm=1\times 10^{-3} m

Force,F=5 N

We have to find the magnitude of the charge on each of these particles.

Let charge on each particle=q

By Coulomb's law of force

F=\frac{kq_1q_2}{r^2}=\frac{kq^2}{r^2}

Where k=9\times 10^9

Substitute the values

5=\frac{9\times 10^9q^2}{(1\times 10^{-3})^2}

q^2=\frac{5\times (1\times 10^{-3})^2}{9\times 10^9}

q=\sqrt{\frac{5\times (1\times 10^{-3})^2}{9\times 10^9}}

q=2.4\times 10^{-8} C

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Answer:

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

Explanation:

The measured pH of the solution = 5.153

C_6H_5OH\rightarrow C_6H_5O^-+H^+

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}

Concentration of phenoxide ions and hydrogen ions are equal to x.

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K_a=9.34\times 10^{-11}

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

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Explanation:

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