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Some Basic Concepts of Chemistry

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Chemistry Part I

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Chemistry

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12

Calculate the mass of 1 a.m.u in grams.


1 space straight a. straight m. straight u space equals space 1 over 12 th space of space the space mass space of space carbon space atom.
space space Now space 6.02 space cross times space 10 to the power of 23 space atoms space of space carbon
space space space space space space space space space space space equals space gram space atomic space mass space of space carbon space equals space 12 space straight g
therefore space space space space Mass space of space one space carbon space atom
space space space space space space space space space space space space equals fraction numerator 12 over denominator 6.02 space cross times space 10 to the power of 23 end fraction space equals space 1.99 space cross times space 10 to the power of negative 23 end exponent
therefore space space space space space 1 space straight a. straight m. straight u. space equals space 1 over 12 cross times 1.99 space cross times space 10 to the power of negative 23 end exponent space equals 1.66 space cross times space 10 to the power of negative 24 end exponent straight g
space space space space space
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In a reaction
straight A plus straight B subscript 2 space rightwards arrow space AB subscript 2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2-5 mol B

(v) 2·5 mol A + 5 mol B


According to the given reaction:

(i) 1 atom of A will react with molecules of B2 = 1
300 atoms of A will react with molecules of B2 = 300
But molecules of B2 actually available = 200
∴  B2 is the limiting reactant.

(ii) 1 mol of A reacts with 1 mol of B
∴ 2 mol of A will react with 2 mol of B.
Hence A is the limiting reactant.

(iii) 100 atoms of A will react with 100 molecules of B. Hence there are no limiting reactants.

(iv) 2·5 mol of B will react with 2·5 mol of A. Hence B is the limiting reagent.

(v) 2·5 mol of A will react with 2·5 mol of B.

Hence A is the limiting reagent.

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The following data were obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen               Mass of dioxygen
(i) 14g                                              16g
(ii) 14g                                             32g
(iii) 28g                                            32g
(iv) 28g                                            80g
which law of chemical combination is obeyed by the above experimental data? Give its statement?


By fixing the mass of dinitrogen as 28g, we can calculate the masses of dioxygen that will combine with 28g of dinitrogn.   
  1 st space compound comma space space space space Mass space of space dioxygen space space equals space 16 over 14 cross times 28 space equals space 32 straight g
2 nd space compound comma space space space Mass space of space dioxygen space space equals space 32 over 14 cross times 28 space equals space 64 space straight g
3 rd space compound space space space space space Mass space of space dioxygen space equals space 32 over 28 cross times 28 space equals space 32 straight g
4 th space compound. space space space Mass space of space dioxygen space space equals 80 over 28 cross times 28 space equals space 80 straight g

These are in the ratio of 32 : 64 : 32 : 80 i.e. 1:2:1:5 which is simple whole number ratio. Hence the given data obeys the law of multiple proportions. For definition refer to the answer of question 40.
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A welding fuel gas contains carbon and hydrogen only. Burning a small space of it in oxygen gives 3.38 g carbon dioxide, 0·690g of water and no other products. A volume of 10·0L (measured at STP) of this welding gas is found to weigh 11·6g. Calculate: (i) empirical formula (ii) molar mass of the gas and (iii) molecular formula.


i.1mole(44g) of CO2 contains 12g of carbon.

3.38 g of CO2 will contain carbon  = (12g/44g) x3.38g =0.9217g

18g of water contains 2g of hydrogen

Therefore 0.690g of water will contain hydrogen=2g/18gx 0.690 =0.0767 g

Since carbon and hydrogen are the only constituents of the compounds, the total mass of the compounds is:

=0.9217+0.767

=0.9984g

Thus,

Percent of C in the compound = (0.9217/.9984) x100 =92.32%

Percent of H in the compound =(0.0767/0.9984) x100=7.68%

Moles of carbon in the compound =92.32/12.00 =7.69

Moles of hydrogen in the compound= 7.68/1=7.68

Since, ration of carbon to hydrogen in the compound=7.69:7.68=1:1

Hence, the empirical formula of the gas is CH.

ii) Given,

Weight of 10.0L of the gas (at S.T.P)=11.6 g

 Therefore, weight of 22.4L of gas at STP  =(11.6/10.0L) x22.4L

     =25.984g

Hence, the molar mass of the gas is 26.0 g.

iii) Empirical formula mass of CH =12+1 =13g

n= molar mass of gas/empirical formula mass of gas

n=26/13

n=2

Therefore, molecular formula of gas= (CH)n­

=C2H2

 

 

299 Views

What do you mean by significant figures?
Or
What do you know about significant figures? What are the rules for determing the number of significant figures?


Significant figures. The total number of digits (including the last one, though its value is doubtful) in the number used to express the physical quantity is called the number of significant figures. For example, the reading 24 · 84 cm having three certain digits (2, 4 and 8) and one doubtful digit (4) has four significant figures.
Rules for determining the number of significant figures:
1. Except zero, all the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are significant. For example,
4·25 has three significant figures.
71·435 has five significant figures.

2. All digits are significant including zeros if zero appears in between non-zero digits, e.g.,
4·02 has three significant figures.
4·005 has four significant figures.
6·0042 has five significant figures.

3. Zeros to the left of the first non-zero digit in a number are not significant, they merely indicate the positions of the decimal point. e.g.,
0·428 has three significant figures.
0·054 has two significant figures.
0·0006 has one significant figure.

4. Zeros to the right of the decimal point are significant. e.g.,
42·0 has three significant figures.
42·00 has four significant figures.
42·000 has five significant figures.

5. In exponential notation, the numerical portion gives the number of significant figures. e.g.,
1·58 × 10–1 has three significant figures.
4·243 × 105 has four significant figures.
3·24 × 107 has three significant figures.

482 Views

Match the following prefixes with their multiples:

A. micro (i) 106
B. deca (ii) 109
C. mega (iii) 10–6
D. giga (iv) 10–15
E. femto (v) 10

A.

micro

(i)

10–6

B.

deca

(ii)

10

C.

mega

(iii)

106

D.

giga

(iv)

109

E.

femto

(v)

10–15
144 Views