19913
According to the first law of Faraday, the weight of a substance discharge at the electrode is
1 \(W = ZQ\)
2 \(W = eF\)
3 \(W = \frac{Z}{F}\,It\)
4 \(W = ZI\)
Explanation:
(a)\(W = ZQ\) ; \(W = Zit\).
ELECTROCHEMISTRY
19914
When \(0.04\) faraday of electricity is passed through a solution of \(CaS{O_4}\), then the weight of \(C{a^{2 + }}\) metal deposited at the cathode is ............. \(\mathrm{gm}\)
19915
A current \( 2.0 \) \(A\) is passed for \(5\) hours through a molten metal salt deposits \(22\,g\) of metal (At. wt. \(=177\)). The oxidation state of the metal in the metal salt is
1 \(+1\)
2 \(+2\)
3 \(+3\)
4 \(+4\)
Explanation:
(c)\({E_{{\rm{metal}}}} = \frac{{{\rm{Weight}}\,\,{\rm{of}}\,{\rm{metal}} \times 96500}}{{{\rm{N}}{\rm{umber}}\,{\rm{of}}\,{\rm{coulombs}}}}\) \( = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} = 59.5\) Oxidation number of the metal \( = \frac{{177}}{{59.5}} = + \,3\)
ELECTROCHEMISTRY
19916
How many atoms of calcium will be deposited from a solution of \(CaC{l_2}\) by a current of \(25\) milliamperes flowing for \(60\) seconds
19913
According to the first law of Faraday, the weight of a substance discharge at the electrode is
1 \(W = ZQ\)
2 \(W = eF\)
3 \(W = \frac{Z}{F}\,It\)
4 \(W = ZI\)
Explanation:
(a)\(W = ZQ\) ; \(W = Zit\).
ELECTROCHEMISTRY
19914
When \(0.04\) faraday of electricity is passed through a solution of \(CaS{O_4}\), then the weight of \(C{a^{2 + }}\) metal deposited at the cathode is ............. \(\mathrm{gm}\)
19915
A current \( 2.0 \) \(A\) is passed for \(5\) hours through a molten metal salt deposits \(22\,g\) of metal (At. wt. \(=177\)). The oxidation state of the metal in the metal salt is
1 \(+1\)
2 \(+2\)
3 \(+3\)
4 \(+4\)
Explanation:
(c)\({E_{{\rm{metal}}}} = \frac{{{\rm{Weight}}\,\,{\rm{of}}\,{\rm{metal}} \times 96500}}{{{\rm{N}}{\rm{umber}}\,{\rm{of}}\,{\rm{coulombs}}}}\) \( = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} = 59.5\) Oxidation number of the metal \( = \frac{{177}}{{59.5}} = + \,3\)
ELECTROCHEMISTRY
19916
How many atoms of calcium will be deposited from a solution of \(CaC{l_2}\) by a current of \(25\) milliamperes flowing for \(60\) seconds
19913
According to the first law of Faraday, the weight of a substance discharge at the electrode is
1 \(W = ZQ\)
2 \(W = eF\)
3 \(W = \frac{Z}{F}\,It\)
4 \(W = ZI\)
Explanation:
(a)\(W = ZQ\) ; \(W = Zit\).
ELECTROCHEMISTRY
19914
When \(0.04\) faraday of electricity is passed through a solution of \(CaS{O_4}\), then the weight of \(C{a^{2 + }}\) metal deposited at the cathode is ............. \(\mathrm{gm}\)
19915
A current \( 2.0 \) \(A\) is passed for \(5\) hours through a molten metal salt deposits \(22\,g\) of metal (At. wt. \(=177\)). The oxidation state of the metal in the metal salt is
1 \(+1\)
2 \(+2\)
3 \(+3\)
4 \(+4\)
Explanation:
(c)\({E_{{\rm{metal}}}} = \frac{{{\rm{Weight}}\,\,{\rm{of}}\,{\rm{metal}} \times 96500}}{{{\rm{N}}{\rm{umber}}\,{\rm{of}}\,{\rm{coulombs}}}}\) \( = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} = 59.5\) Oxidation number of the metal \( = \frac{{177}}{{59.5}} = + \,3\)
ELECTROCHEMISTRY
19916
How many atoms of calcium will be deposited from a solution of \(CaC{l_2}\) by a current of \(25\) milliamperes flowing for \(60\) seconds
NEET Test Series from KOTA - 10 Papers In MS WORD
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ELECTROCHEMISTRY
19913
According to the first law of Faraday, the weight of a substance discharge at the electrode is
1 \(W = ZQ\)
2 \(W = eF\)
3 \(W = \frac{Z}{F}\,It\)
4 \(W = ZI\)
Explanation:
(a)\(W = ZQ\) ; \(W = Zit\).
ELECTROCHEMISTRY
19914
When \(0.04\) faraday of electricity is passed through a solution of \(CaS{O_4}\), then the weight of \(C{a^{2 + }}\) metal deposited at the cathode is ............. \(\mathrm{gm}\)
19915
A current \( 2.0 \) \(A\) is passed for \(5\) hours through a molten metal salt deposits \(22\,g\) of metal (At. wt. \(=177\)). The oxidation state of the metal in the metal salt is
1 \(+1\)
2 \(+2\)
3 \(+3\)
4 \(+4\)
Explanation:
(c)\({E_{{\rm{metal}}}} = \frac{{{\rm{Weight}}\,\,{\rm{of}}\,{\rm{metal}} \times 96500}}{{{\rm{N}}{\rm{umber}}\,{\rm{of}}\,{\rm{coulombs}}}}\) \( = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} = 59.5\) Oxidation number of the metal \( = \frac{{177}}{{59.5}} = + \,3\)
ELECTROCHEMISTRY
19916
How many atoms of calcium will be deposited from a solution of \(CaC{l_2}\) by a current of \(25\) milliamperes flowing for \(60\) seconds
19913
According to the first law of Faraday, the weight of a substance discharge at the electrode is
1 \(W = ZQ\)
2 \(W = eF\)
3 \(W = \frac{Z}{F}\,It\)
4 \(W = ZI\)
Explanation:
(a)\(W = ZQ\) ; \(W = Zit\).
ELECTROCHEMISTRY
19914
When \(0.04\) faraday of electricity is passed through a solution of \(CaS{O_4}\), then the weight of \(C{a^{2 + }}\) metal deposited at the cathode is ............. \(\mathrm{gm}\)
19915
A current \( 2.0 \) \(A\) is passed for \(5\) hours through a molten metal salt deposits \(22\,g\) of metal (At. wt. \(=177\)). The oxidation state of the metal in the metal salt is
1 \(+1\)
2 \(+2\)
3 \(+3\)
4 \(+4\)
Explanation:
(c)\({E_{{\rm{metal}}}} = \frac{{{\rm{Weight}}\,\,{\rm{of}}\,{\rm{metal}} \times 96500}}{{{\rm{N}}{\rm{umber}}\,{\rm{of}}\,{\rm{coulombs}}}}\) \( = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} = 59.5\) Oxidation number of the metal \( = \frac{{177}}{{59.5}} = + \,3\)
ELECTROCHEMISTRY
19916
How many atoms of calcium will be deposited from a solution of \(CaC{l_2}\) by a current of \(25\) milliamperes flowing for \(60\) seconds