For exhibiting cis-trans (or geometrical isomerism, a molecule must fulfil the following condition.
The groups attached to each double bonded carbon atom must be different.

Percentage of carbon =\(\frac { 12\times 0.198\times 100 }{ 44\times 0.246 } =21.95%\) %
Percentage of hydrogen =\(\frac { 2\times 0.1014\times 100 }{ 18\times 0.246 } =4.58%\) %

The basic units for length is meter (m), for time is second(s) and for mass is kilogram (kg)
\(15.15\mu s\frac { { 10 }^{ -12 }s }{ 1\mu s } =1.515\times { 10 }^{ -5 }s\)

\(\text{ we know } \triangle _{ r }G^{ \ominus }=-2.303 \ RT \ \log { { K }_{ P } }\) \(and R=8.314 \ JK^{ -1 }mol^{ -1 }\)
\(\text{ therefore,}\triangle _{ r }G^{ \ominus }=-2.303(831JK^{ -1 }mol^{ -1 })\times (298K)(\log { 2.47\times { 10 }^{ -29 }) }\)
= 163000 J mol^–1
= 163 kJ mol^–1.

Molar mass of AgBr = 108 + 80 = 188 g mol^-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains \(= \frac{80 \times 0.12}{188}\) g bromine
Percentage of bromine \(=\frac { 80\times 0.12\times 100 }{ 188\times 0.15 } =34.04\%\)

M = 3 mol L^-1
Mass of NaCl in 1 L soution = x 58.5 = 175.5 g
Mass of 1 L solution =1000 x 1.25 = 1250 g
(since density = 1.25 g mL^-1)
Mass of water soultion = 1250 - 175.5
= 1074.5 g = 1.0745 kg.
Molality = \(\frac { \text{ number of moles of solute} }{\text{ mass of solvent in kg} } =\frac { 3mol }{ 1.0745kg } =2.79 \ m\)
Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.

Given, (5.7 x 10^-5) \(\div \) (4.2 x 10^-3)
(5.7 \(\div \) 4.2) x (10^-5-(-3)) = 23.94 x 10^-2

LiCl is soluble in organic solvent but KCl does not. THis is because exceptionaloly small size of Li⁺ ion and high polarising power, which develops covalent character in the lithium compounds ( Fajan's Rule ) .whereas , KCl is ionic in nature .So, it is soluble in water.

Milk, Yogurt and milk products are rich sources of calcium

The value possessed by the office bearer of Green Park Association are best to use of rainwater, saving water for drinking irrigation, domestic purpose etc.

Among the oxoanions of chlorine listed above, \(Cl{ O }_{ 4 }^{ - }\)  does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine are as follows:
\(\overset { +1 }{ 3ClO^{ - } } \longrightarrow \overset { -1 }{ 2Cl^{ - } } +\overset { +5 }{ ClO_{ 3 }^{ - } }\)
\( \overset { +3 }{ 6Cl{ O }_{ 2 }^{ - } } \overset { hv }{ \longrightarrow } \overset { +5 }{ 4Cl{ O }_{ 3 }^{ - }+ } \overset { -1 }{ 2C } l^{ - }\)
\(4\overset { +5 }{ ClO_{ 3 }^{ - } \longrightarrow \overset { { -1 }^{ - } }{ Cl } } +3\overset { +7 }{ Cl } _{ 4 }^{ - }\)

Step I :  To calculate empirical formula of the compound.
Here,  % of O = 100 - ( 40.687 + 5.085 )
= 54.228%
\(\therefore\) Empirical formula is C₂HO₂
Step II :   To calculate empirical formula mass. The empirical formula of the compound is C2H3O2.
\(\therefore\)  Empirical formula mass = ( 2\(\times\)12 ) + ( 3\(\times\)1 ) + ( 2\(\times\)16 )
= 59
Step III : To calculate the molecular mass of the salt.The vapour density of the compound  = 59 ( Given )
Using the relation between vapour density and molecular mass.
We have molecular mass = 2 \(\times\) vapour density
= 2\(\times\)59 = 118
Step IV : To calculate the value of n = \(\frac{molecular \ mass}{emphirical \ formula \ mass}\)
= \(\frac{118}{59}\) 
= 2 
Step V : To calculate the molecular formula of the salt , Molecular formula = n \(\times\) emphirical formula
= 2C₂H₃O₂
= C₄H₆O₄
Thus, the molecular formula is C₄H₆O₄.

1 mole of H2O = 18 cm3
( \(\because\)  density of H₂O = 1g / cm³)
= 6.022\(\times\)10²³ molecules of H₂O
Thus, 6.022\(\times\)10²³ molecules of H₂O have volume
= 18 cm³
\(\therefore\) 1 molecule of H₂O will have volume
= \(\frac{18}{6.022\times10^{23}} \)
= 2.989\(\times\)10^-23 cm³