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2 years ago

Which of the following is most likely to form multiple (double or triple) bonds? 1. Li 2. Cl 3. N 4. H 5. F

Chemistry

1 answer:

alukav5142 [94]2 years ago

3 0

Answer: Option (3) is the correct answer.

Explanation:

Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.

Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.

Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.

Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.

Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.

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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

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Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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Read 2 more answers

β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa

gtnhenbr [62]

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273.15 +25]=298.15 K$

Putting values in above equation, we get:

$0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L$

The concentration of solute is $2.4278\times 10^{-5} mol/L$

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

$C=\frac{n}{V(L)}$

$n=2.4278\times 10^{-5} mol/L\times 0.137 L$

$n=3.3261\times 10^{-6} mol$

To calculate the molecular mass of solute, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of β‑Galactosidase = $3.3261\times 10^{-6} mol$

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

$3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol$

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

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