If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
Diameter of carbon atom = 0.15nm = 0.15 x 10-9m = 1.5 x 10-10m
Length along which atoms are to be placed = 20cm = 20x10-2m = 2 x 10-1m
Therefore no of carbon atoms which can be placed along the length
= (2 x 10-1m) / (1.5x10-10 )
=1.33x109
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Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
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(a) n = 1, l = 0;
(b) n = 3; l =1
(c) n = 4; l = 2;
(d) n = 4; l =3.
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Na+, K+, Mg2+, Ca2+, S2–, Ar
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(a) n = 4,
(b) n = 3, l = 0
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(iv) CH2=CHCN,
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(ii) CO2
(iii) CH4
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(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
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(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
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(a)
(b)
(c)
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(i) CH2=C=O,
(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
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