Chemistry Questions and Answers

No explanation has been provided for this answer.

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