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3 years ago

As atomic size increases, how do electronegativity and ionization energy change?

1 answer:

3 0

Ionisation of energy is the amount of energy needed to remove an electron from outer most shell of an atom. as the attraction between the nucleus and outer most electron increases the ionisation energy increases. Electronegativity is the relative measure of the force of attraction between an atom and the electron it is sharing in bond with other atom. electronegativity decreases as it goes down the group.

In conclusion:

As an atom gain an electron the atomic radius of the atom increase and the ionisation energy also increases but the electronegativity decrease as the atoms in outer most shell move further apart from the nucleus.

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The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

svlad2 [7]

Answer: Dissociation constant of the acid is $10^{-11}$ .

Explanation: Assuming the acid to be monoprotic, the reaction follows:

$HA\rightleftharpoons H^++A^-$

pH of the solution = 6

and we know that

$pH=-log([H^+])$

$[H^+]=antilog(-pH)$

$[H^+]=antilog(-6)=10^{-6}M$

As HA ionizes into its ions in 1 : 1 ratio, hence

$[H^+]=[A^-]=10^{-6}M$

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

$[HA]=[HA]-[H^+]$

$[HA]=0.1M-10^{-6}M$

$[HA]=0.09999M$

Dissociation Constant of acid, $K_a$ is given as:

$K_a=\frac{[A^-][H^+]}{HA}$

Putting values of $[H^+],[A^-]\text{ and }[HA]$ in the above equation, we get

$K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}$

$K_a=1.0001\times 10^{-11}M$

Rounding it of to one significant figure, we get

$K_a=1.0\times 10^{-11}M\approx 10^{-11}M$

6 0

3 years ago

Difne substitution reaction with example

alisha [4.7K]

Answer:

A substitution reaction is also called a single displacement reaction, single replacement reaction, or single substitution reaction. Examples: CH3Cl reacted with a hydroxy ion (OH-) will produce CH3OH and chlorine. This substitution reaction replaces the chlorine atom on the original molecule with the hydroxy ion.

Hope it helps!!!

5 0

3 years ago

Read 2 more answers

The constant- pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression

vivado [14]

A perfect gas, commonly known as an ideal gas, would be a gas that, in physical behavior, conforms towards the general gas law, which would be a <u><em>specific, idealized relationship between pressure, volume, and temperature,</em></u> and further calculation can be defined as follows:

For question a:

$q=\int C_p dT=\int^{200+273K}_{25+273 K} [20.17 + 0.3665 (\frac{T}{K})] dT JK^{-1}\\\\$

$=[(20.17)T +\frac{1}{2}(0.3665)\times (\frac{T^2}{K})]^{473K}_{298K}\ JK^{-1}\\\\=[(20.17)\times(473-293) +\frac{1}{2}(0.3665)\times (473^2-293^2)]\ J\\\\=[(20.17)\times(180) +\frac{1}{2}(0.3665)\times (223729-85849)]\ J\\\\=[3630.6 +(0.18325)\times (137880)]\ J\\\\=[3630.6 +25266.51]\ J\\\\=[28897.11]\ J\\\\=[28.8 \times 10^{3}] \ J\\\\$

$W= -p \Delta V= -nR\Delta T= -(1.00\ mol) \times 8.3145\ J K^{-1}\ mol^{-1} \times 100\ K=-831\ J\\\\\Delta U= q+w=(28.8 -0.831)\ KJ= 27.969\ KJ\\\\$

For question b:

In this question, the energy and enthalpy of a perfect gas depend on temperature alone, therefore $\Delta H= 28.8 \ KJ$ and $\Delta U= 27.969\ KJ$ as above at constant volume $w=0$ and $\Delta U=q$ so, $q=27.969 \ KJ$ .