Al\(^{3+}\), Cl\(^-\)
Hg\(^+\), SO\(_4 ^{2-}\)
Pb\(^{2+}\), Cl\(^-\)
Ag\(^+\), SO\(_4 ^{2-}\)
Correct answer is C
No explanation has been provided for this answer.
The reaction \(2H_2 S + SO_2 \to 3S + 2H_2 O\) is
not a redox reaction because there is no oxidant in the reaction
not a redox reaction because there is no reductant in the reaction
a redox reaction in which H\(_2\)S is the oxidant and SO\(_2\) is the reductant
a redox reaction in which SO\(_2\) is the oxidant and H\(_2\)S is the reductant
Correct answer is D
No explanation has been provided for this answer.
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