37512
The volume of \(\frac{N}{{10}}\) \(NaOH\) require to neutralise \(100\, ml\) of \(\frac{N}{{25}}\) \(HCl\) is......\(ml\)
1 \(30\)
2 \(100\)
3 \(40\)
4 \(25\)
Explanation:
(c) For complete neutralisation, milli equivalent of base $=$ milli equivalent of acid \({N_1}{V_1} = {N_2}{V_2}\) $⇒$ \(\frac{1}{{10}} \times {V_1} = \frac{1}{{25}} \times 100\); \({V_1} = 40\;ml.\)
PRACTICAL CHEMISTRY
37513
The volume of \(0.6\, M\) \(NaOH\) required to neutralise \(30\) \(c{m^3}\) of \(0.4 \,M\) \(HCl\) is........\(c{m^3}\)
37515
If \(30 \,ml\) of \({H_2}\) and \(20 \,ml\) of \({O_2}\) reacts to form water, what is left at the end of the reaction
1 \(10\,ml\) of \({H_2}\)
2 \(5\,ml\) of \({H_2}\)
3 \(10\,ml \) of \({O_2}\)
4 \(5\,ml \) of \({O_2}\)
Explanation:
(d) \({H_2} + \frac{1}{2}{O_2} \to {H_2}O\) \(1\) \(mole\) \(\frac{1}{2}\) \(mole\) \(1\) \(mole\) \(1\) volume \(\frac{1}{2}\) volume \(1\, ml\) \({H_2}\) reacts with \(\frac{1}{2}\) \(ml\) \({O_2}\) \(30\, ml\) of \({H_2}\) reacts with \( = \frac{1}{2} \times 30 = 15\,ml\) \({O_2}\) \((20-15) = 5\,ml\) of \({O_2}\) will left at the end of the reaction.
37512
The volume of \(\frac{N}{{10}}\) \(NaOH\) require to neutralise \(100\, ml\) of \(\frac{N}{{25}}\) \(HCl\) is......\(ml\)
1 \(30\)
2 \(100\)
3 \(40\)
4 \(25\)
Explanation:
(c) For complete neutralisation, milli equivalent of base $=$ milli equivalent of acid \({N_1}{V_1} = {N_2}{V_2}\) $⇒$ \(\frac{1}{{10}} \times {V_1} = \frac{1}{{25}} \times 100\); \({V_1} = 40\;ml.\)
PRACTICAL CHEMISTRY
37513
The volume of \(0.6\, M\) \(NaOH\) required to neutralise \(30\) \(c{m^3}\) of \(0.4 \,M\) \(HCl\) is........\(c{m^3}\)
37515
If \(30 \,ml\) of \({H_2}\) and \(20 \,ml\) of \({O_2}\) reacts to form water, what is left at the end of the reaction
1 \(10\,ml\) of \({H_2}\)
2 \(5\,ml\) of \({H_2}\)
3 \(10\,ml \) of \({O_2}\)
4 \(5\,ml \) of \({O_2}\)
Explanation:
(d) \({H_2} + \frac{1}{2}{O_2} \to {H_2}O\) \(1\) \(mole\) \(\frac{1}{2}\) \(mole\) \(1\) \(mole\) \(1\) volume \(\frac{1}{2}\) volume \(1\, ml\) \({H_2}\) reacts with \(\frac{1}{2}\) \(ml\) \({O_2}\) \(30\, ml\) of \({H_2}\) reacts with \( = \frac{1}{2} \times 30 = 15\,ml\) \({O_2}\) \((20-15) = 5\,ml\) of \({O_2}\) will left at the end of the reaction.
37512
The volume of \(\frac{N}{{10}}\) \(NaOH\) require to neutralise \(100\, ml\) of \(\frac{N}{{25}}\) \(HCl\) is......\(ml\)
1 \(30\)
2 \(100\)
3 \(40\)
4 \(25\)
Explanation:
(c) For complete neutralisation, milli equivalent of base $=$ milli equivalent of acid \({N_1}{V_1} = {N_2}{V_2}\) $⇒$ \(\frac{1}{{10}} \times {V_1} = \frac{1}{{25}} \times 100\); \({V_1} = 40\;ml.\)
PRACTICAL CHEMISTRY
37513
The volume of \(0.6\, M\) \(NaOH\) required to neutralise \(30\) \(c{m^3}\) of \(0.4 \,M\) \(HCl\) is........\(c{m^3}\)
37515
If \(30 \,ml\) of \({H_2}\) and \(20 \,ml\) of \({O_2}\) reacts to form water, what is left at the end of the reaction
1 \(10\,ml\) of \({H_2}\)
2 \(5\,ml\) of \({H_2}\)
3 \(10\,ml \) of \({O_2}\)
4 \(5\,ml \) of \({O_2}\)
Explanation:
(d) \({H_2} + \frac{1}{2}{O_2} \to {H_2}O\) \(1\) \(mole\) \(\frac{1}{2}\) \(mole\) \(1\) \(mole\) \(1\) volume \(\frac{1}{2}\) volume \(1\, ml\) \({H_2}\) reacts with \(\frac{1}{2}\) \(ml\) \({O_2}\) \(30\, ml\) of \({H_2}\) reacts with \( = \frac{1}{2} \times 30 = 15\,ml\) \({O_2}\) \((20-15) = 5\,ml\) of \({O_2}\) will left at the end of the reaction.
37512
The volume of \(\frac{N}{{10}}\) \(NaOH\) require to neutralise \(100\, ml\) of \(\frac{N}{{25}}\) \(HCl\) is......\(ml\)
1 \(30\)
2 \(100\)
3 \(40\)
4 \(25\)
Explanation:
(c) For complete neutralisation, milli equivalent of base $=$ milli equivalent of acid \({N_1}{V_1} = {N_2}{V_2}\) $⇒$ \(\frac{1}{{10}} \times {V_1} = \frac{1}{{25}} \times 100\); \({V_1} = 40\;ml.\)
PRACTICAL CHEMISTRY
37513
The volume of \(0.6\, M\) \(NaOH\) required to neutralise \(30\) \(c{m^3}\) of \(0.4 \,M\) \(HCl\) is........\(c{m^3}\)
37515
If \(30 \,ml\) of \({H_2}\) and \(20 \,ml\) of \({O_2}\) reacts to form water, what is left at the end of the reaction
1 \(10\,ml\) of \({H_2}\)
2 \(5\,ml\) of \({H_2}\)
3 \(10\,ml \) of \({O_2}\)
4 \(5\,ml \) of \({O_2}\)
Explanation:
(d) \({H_2} + \frac{1}{2}{O_2} \to {H_2}O\) \(1\) \(mole\) \(\frac{1}{2}\) \(mole\) \(1\) \(mole\) \(1\) volume \(\frac{1}{2}\) volume \(1\, ml\) \({H_2}\) reacts with \(\frac{1}{2}\) \(ml\) \({O_2}\) \(30\, ml\) of \({H_2}\) reacts with \( = \frac{1}{2} \times 30 = 15\,ml\) \({O_2}\) \((20-15) = 5\,ml\) of \({O_2}\) will left at the end of the reaction.
37512
The volume of \(\frac{N}{{10}}\) \(NaOH\) require to neutralise \(100\, ml\) of \(\frac{N}{{25}}\) \(HCl\) is......\(ml\)
1 \(30\)
2 \(100\)
3 \(40\)
4 \(25\)
Explanation:
(c) For complete neutralisation, milli equivalent of base $=$ milli equivalent of acid \({N_1}{V_1} = {N_2}{V_2}\) $⇒$ \(\frac{1}{{10}} \times {V_1} = \frac{1}{{25}} \times 100\); \({V_1} = 40\;ml.\)
PRACTICAL CHEMISTRY
37513
The volume of \(0.6\, M\) \(NaOH\) required to neutralise \(30\) \(c{m^3}\) of \(0.4 \,M\) \(HCl\) is........\(c{m^3}\)
37515
If \(30 \,ml\) of \({H_2}\) and \(20 \,ml\) of \({O_2}\) reacts to form water, what is left at the end of the reaction
1 \(10\,ml\) of \({H_2}\)
2 \(5\,ml\) of \({H_2}\)
3 \(10\,ml \) of \({O_2}\)
4 \(5\,ml \) of \({O_2}\)
Explanation:
(d) \({H_2} + \frac{1}{2}{O_2} \to {H_2}O\) \(1\) \(mole\) \(\frac{1}{2}\) \(mole\) \(1\) \(mole\) \(1\) volume \(\frac{1}{2}\) volume \(1\, ml\) \({H_2}\) reacts with \(\frac{1}{2}\) \(ml\) \({O_2}\) \(30\, ml\) of \({H_2}\) reacts with \( = \frac{1}{2} \times 30 = 15\,ml\) \({O_2}\) \((20-15) = 5\,ml\) of \({O_2}\) will left at the end of the reaction.
