How many alkoxyalkanes can be obtained from the molecular formula C\(_4\)H\(_{10}\)O?
4
2
1
3
Correct answer is D
No explanation has been provided for this answer.
N12.00
N27.00
N63.00
N95.00
Correct answer is C
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
N10.00
N6.00
N20.00
N9.00
Correct answer is D
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
Hydrogen diffused through a porous plug
thrice as fast as oxygen
at the same speed as oxygen
four times as fast as oxygen
twice as fast as oxygen
Correct answer is C
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.
The emission of two successive beta particles from the nucleus \(^{32} _{15} P\) will produce
\(^{30} _{14}\)Si
\(^{32} _{13}\)Al
\(^{32} _{17}\)Cl
\(^{32} _{16}\)S
Correct answer is C
\(^{32} _{15}\)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