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tiny-mole [99]
3 years ago
5

A mixture of helium, nitrogen and oxygen has a total pressure of 756 mmHg. The partial

Chemistry
1 answer:
Karolina [17]3 years ago
6 0

Answer:

The  partial pressure of oxygen in the mixture is 296 mmHg.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone.

This relationship is due to the assumption that there are no attractive forces between the gases.

So, in this case, the total pressure is:

PT=Phelium + Pnitrogen + Poxygen

You know:

  • PT= 756 mmHg
  • Phelium= 122 mmHg
  • Pnitrogen= 338 mmHg
  • Poxygen= ?

Replacing:

756 mmHg= 122 mmHg + 338 mmHg + Poxygen

Solving:

756 mmHg - 122 mmHg - 338 mmHg = Poxygen

Poxygen= 296 mmHg

<u><em>The  partial pressure of oxygen in the mixture is 296 mmHg.</em></u>

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In addition to mathematics, physics and astronomy, Newton also had an interest in alchemy, mysticism, and theology.

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What mass of HBO2 is produced from the<br> combustion of 139.5 g of B2H6?<br> Answer in units of g.
aliina [53]

Answer:

m_{HBO_2}=441.8gHBO_2

Explanation:

Hello there!

In this case, since the combustion of B2H6 is:

B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O

Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:

m_{HBO_2}=139.5gB_2H_6*\frac{1molB_2H_6}{27.67gB_2H_6} *\frac{2molHBO_2}{1molB_2H_6} *\frac{43.82gHBO_2}{1molHBO_2}

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2 years ago
In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}

n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{Fe} = 0.5\:mol}

* to n(S)

n_{S} = \dfrac{m_{S}}{MM_{S}}

n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{S} = 0.5\:mol}

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

Fe + S \Longrightarrow FeS

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}

m_{FeS} = n_{FeS}*MM_{FeS}

m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

44 grams of iron sulfide

___________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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Ask a question.

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Explanation:

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