Chemistry

Kindly consider the following figure

Probability density of plots

From the given graph answer is (1)

Hess's law
ΔH_solution = ΔH_lattice + ΔH_hydration
4 = 788 + ΔH_hydration
ΔH_hydration = -784 kJ mol? ¹

In isoelectronic species nuclear charge can be approximated as
Nuclear charge ≈ z / no. of electrons
Al³? Mg²? Na? F? O²? N³?
Nuclear Charge: 13/10 12/10 11/10 9/10 8/10 7/10
Minimum nuclear charge is in N³? and maximum is in Al³?
So order should be
Al³? < Mg? < Na? < F? < O? < N?

Using VSEPR, L.P - B.P repulsion we can safely say that CH? should have the highest bond angle among the given

KNO? , HCl and NaCl are strong electrolytes for these electrolytes of Λ_m with √c will be liner which can be given as
Λ_m = Λ? _m - A√c for strong electrolyte
Since given variation is not linear it has to be a weak electrolyte
CH? COOH is a weak electrolyte

Gas + Solid? GS? H = -ve
Adsorbed gas
Adsorption of gas is an exothermic process. Increase in temperature reduces the extent of adsorption.
x/m = Kp¹/? (n > 1)

$\mathrm{}{f}^{\mathrm{\text{'}}}\left(x\right)=x\left(\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?(\mathrm{s}\mathrm{i}\mathrm{n}?|x\left|\right)\right)=x\left(\pi -\left(\frac{\pi }{2}-{\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?(\mathrm{s}\mathrm{i}\mathrm{n}?|x\left|\right)\right)\right)=x\left(\frac{\pi }{2}+\left|x\right|\right)$

$f\left(x\right)=\left\{\begin{array}{ll}x\left(\frac{\pi }{2}+x\right)& x\ge 0\\ x\left(\frac{\pi }{2}-x\right)& x<0\end{array}\right.$

$f^{\mathrm{\text{'}}}\left(x\right)=\left\{\begin{array}{ll}\frac{\pi }{2}+2x& x\ge 0\\ \frac{\pi }{2}-2x& x<0\end{array}\right.$  is increasing in $f^{\mathrm{\text{'}}}\left(x\right)$ ${}^{}$ and decreasing in $\left(0,\frac{\pi }{2}\right)$

In FCC octahedral voids are present at the edge centers and body center

Consider a diagonal projected form edge centre passing through the body centre
Distance between octahedral voids = √2a/2 = a/√2

$\mathrm{}496.00$

$\mathrm{}PPM=\frac{\text{Mass of Iron}}{\text{Mass of wheat}}*{10}^6$

Mass of $10=\frac{\text{Mass of Iron}}{100*{10}^3}*{10}^6$ $\frac{}{{}^{}}{}^{}$

Mole of $\mathrm{F}\mathrm{e}=1\mathrm{g}\mathrm{m}$ $$

$$ $\mathrm{F}\mathrm{e}=\frac{1}{56}$  contain ${\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{O}}_{4}7{\mathrm{H}}_2\mathrm{O}$ ${}_{}{}_{}$ mole of $$ atom $=1$

$$  $\mathrm{F}\mathrm{e}$ in 1 mole

$\therefore 56\mathrm{g}$  in  $1\mathrm{g}$ $$ mole

$\frac{}{}$  $\frac{1}{56}$ Mass $\therefore$ $$