Number of valency electrons in 4.2gram of \(N_{3}^{-}\)ion is
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Solution
Number of valence electrons in a \(N_{3}^{-}\)ion = 1
Now, 1 mol or 42 g of \(N_{3}^{-}\)has = 6.023 × 1023 ions
So, 42 g of \(N_{3}^{-}\) has 6.023 × 4 × 1023 valency e-
1 g of \(N_{3}^{-}\) has\(\frac{6.023\times 1\times 10^{23}}{42}\)valency e-
4.2 g of \(N_{3}^{-}\) has\(\frac{4.2\times 6.023\times 1\times 10^{23}}{42}\) valency e– i.e., 0.1 NA valency e–.
Two samples of lead oxide were separately reduced to metallic lead by heating in acurrent of hydrogen.The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates
The prefix zepto stands for (in m)
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Solution
1 zepto =10-21
Given P = 0.0030m, Q = 2.40m, R = 3000m, Significant figures in P, Q and R are respectively
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Solution
Given P =0.0030m, Q = 2.40m & R = 3000m. In P(0.0030)initial zeros after the decimal point are not significant.Therefore, significant figures in P(0.0030) are 2. Similarly in Q (2.40)significant figures are 3 as in this case final zero is significant.In R = (3000)all the zeros are significant hence, in R significant figures are 4 because they come from a measurement.
The volume of 20 volume H2O2 required to get 5 litres of O2 at STP is
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Solution
20 volume H2O2 means that 1mL of this H2O2 solutions produces 20 mL of O2 at N.T.P.on decomposition by heat.
∴For 20 mL of O2, the volume of 20 volume H2O2 required = 1mL For 1mL of O2, the volume of 20 volume H2O2 required = 1⁄20 mL
For 5000 mL or 5L of O2, the volume of 20
volume H2O2 required =1⁄20×5000mL = 250 mL
Irrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
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Solution
The H : O ratio in water is fixed, irrespective of its source.Hence it is law of constant composition.
Among the following pairs of compounds, the one that illustrates the law of multiple proportions is
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Solution
In CuO and Cu2O the O :Cu is1 : 1 and 1 : 2 respectively.This is law of multiple proportion
The weight of one molecule of a compound of molecular formula C60H122is
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Solution
M.W. = 60 × 12+ 122 =842
Weight of one molecule =\(\frac{842}{6.02\times 10^{23}}\)gm
= 140 × 10–23gm = 1.4 × 10-21gm
A sample was weighted using two different balances. The results were
(i)3.929 g(ii)4.0 g
How would the weight of the sample be reported?
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Solution
Out of two 3.929 g is more accurate and will be reported as 3.93 after rounding off.
The prefix 1018 is
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Solution
Exa = 1018