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3 years ago

Calculate the percent covalent character of a chemical bond between oxygen and beryllium.

Chemistry

1 answer:

dezoksy [38]3 years ago

8 0

Beryllium Oxide has a molecular formula Be-O

% covalent character in Be-O = 100 - % ionic character in Be-O

Now,

% ionic character = {1 - exp[(-0.25)(E(Be)-E(O))²]}*100

where,

E(Be) and E(O) are the electronegativity values of Be and O

Based on the Paulings scale:-

E(Be) = 1.57

E(O) = 3.44

% ionic = {1 - exp[(-0.25)(1.57-3.44)²]}*100 = 58.28%

Therefore, % covalent = 100 -58.28 = 41.72%

Ans: b) 41 percent

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Marat540 [252]

Answer : The partial pressure of helium is, $1.815\times 10^4KPa$

Solution : Given,

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$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{10g}{32g/mole}=0.3125moles$

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{50g}{4g/mole}=12.5moles$

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{40g}{28g/mole}=1.428moles$

Now we have to calculate the total number of moles of gas mixture.

$\text{Total number of moles of gas}=\text{Moles of oxygen gas}+\text{Mole of helium gas}+\text{Moles of nitrogen gas}$

$\text{Total number of moles of gas}=0.3125+12.5+1.428=14.24moles$

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$\text{Mole fraction of He gas}=\frac{\text{Moles of He gas}}{\text{Total number of moles of gas}}=\frac{12.5}{14.25}=0.877$

Now we have to calculate the partial pressure of helium.

$p_{He}=X_{He}\times P_T$

where,

$p_{He}$ = partial pressure of helium

$P_T$ = total pressure

$X_{He}$ = mole fraction of helium

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$p_{He}=(0.877)\times (2.07\times 10^4KPa)=1.815\times 10^4KPa$

Therefore, the partial pressure of helium is, $1.815\times 10^4KPa$

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