Chemistry Questions and Answers

For X:  X\(^{2+}\) + 2e\(^-\) \(\to\) X

2F = 63g

xF = 6g

x = \(\frac{6 \times 2}{63} = \frac{4}{21} F\)

\(\frac{4}{21}\)F = N12.00

1F = \(\frac{12}{\frac{4}{21}}\)

= N63.00

1F is equivalent to N63.00.

For Y: Y\(^{3+}\) + 3e\(^-\) \(\to\) Y

3F = 27g 

xF = 9g

x = \(\frac{3 \times 9}{27}\)

= 1F

1F = N63.00

X\(^{3+}\) + 3e\(^-\) \(\to\) X

3F = 27g

xF = 10g

\(\frac{x}{3} = \frac{10}{27} \implies x = \frac{10}{9} F\)

\(\frac{10}{9}\)F \(\equiv\) N20.00

1F is equivalent to x

\(\frac{1}{\frac{10}{9}} = \frac{x}{20}\)

\(\frac{9}{10} = \frac{x}{20} \implies x = N18.00\)

1F is equivalent to N18.00.

Y\(^{2+}\) + 2e\(^-\) \(\to\) Y

2F = 24g

xF = 6g

x = \(\frac{6 \times 2}{24} = \frac{1}{2} F\)

1F = N18.00

\(\frac{1}{2}\)F = \(\frac{1}{2} \times N18.00\)

= N9.00

Using Graham's Law of Diffusion,

\(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\)

\(\frac{R_H}{R_O} = \sqrt{16} = 4\)

\(\therefore R_H = 4 \times R_O\)

Therefore, hydrogen diffuses 4 times faster than oxygen.

\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X

32 = 2(0) + a

a = 32

15 = 2(-1) + b

b = 15 + 2

b = 17

\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl

Recall:

V = \(\sqrt{\frac{3RT}{M}}\)

\(\therefore V \propto \frac{1}{\sqrt{M}}\)

\(V = \frac{k}{\sqrt{M}}\)

V = \(\frac{k}{M^{\frac{1}{2}}}\)