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3 years ago

What is the isotopic notation for phosphorus

Chemistry

1 answer:

den301095 [7]3 years ago

3 0

Answer:

Name Phosphorus

Symbol P

Atomic Number 15

Atomic Mass 30.974 atomic mass units

Number of Protons 15

Explanation:

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2 years ago

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Answer:

For 1: The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2: The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

Explanation:

For 1:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For helium:

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

For methane gas:

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

For Helium gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

For Methane gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2:

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

For nitrogen gas:

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

For carbon dioxide gas:

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

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