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3 years ago

What mass of xenon tetrafluoride, xef 4 , has the same number of fluorine atoms as 25.0 g of oxygen difluoride, of2?

2 answers:

6 0

To determine the mass of xenon tetrafluoride, we need to know first the number of fluorine atoms present in oxygen difluoride. We need to convert first the mass into moles then make use of the relation of the elements from the chemical formula. Then, use the avogadro's number to convert it to number of atoms. Then, we do the reverse of the steps above but this time for xenon tetrafluoride.

25.0 g OF2 ( 1 mol / 54 g ) ( 2 mol F / 1 mol OF2 ) ( 6.022 x10^23 atoms F / 1 mol F ) ( 1 mol / 6.022x10^23 atoms) ( 1 mol XeF4 / 4 mol F ) (207.3 g / 1 mol XeF4) = 47.99 g XeF4

kati45 [8]3 years ago

6 0

Answer is: mass of xenon tetrafluoride is 47.98 grams.

1) m(OF₂) = 25.0 g; mass of oxygen difluoride.

n(OF₂) = m(OF₂) ÷ M(OF₂).

n(OF₂) = 25 g ÷ 54 g/mol.

n(OF₂) = 0.462 mol; amount of substance.

In one oxygen difluoride (OF₂) there are two fluorine atoms: n(F) : n(OF₂) = 2 : 1.

n(F) = 0.462 mol · 2.

n(F) = 0.926 mol.

N(F) = n(F) · Na(Avogadro constant).

N(F) = 0.926 mol · 6.022·10²³ 1/mol.

N(F) = 5.576·10²³; number of fluorine atoms in oxygen difluoride.

2) In one molecule of xenon tetrafluoride there are four fluorine atoms:

n(XeF₄) : n(F) = 1 :4.

n(XeF₄) = 0.926 mol ÷ 4.

n(XeF₄) = 0.2315 mol; amount of xenon tetrafluoride.

m(XeF₄) = 0.2315 mol · 207.28 g/mol.

m(XeF₄) = 47.98 g.

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myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

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Explanation:

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Answer: C. no new substances are formed

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Answer:

A. there is an isotope of lanthanum with an atomic mass of 138.9

Explanation:

By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.

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Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.

Thus, the correct option is:

A. there is an isotope of lanthanum with an atomic mass of 138.9.

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