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3 years ago

A solution of hydrochloric acid contains 3.2 g of hydrogen chloride in 50 cm3

Chemistry

1 answer:

Harrizon [31]3 years ago

4 0

The concentration of Hydrogen chloride in g per dm³ = 64 g/dm³

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent

A solution of hydrochloric acid contains 3.2 g of hydrogen chloride in 50 cm³, so the concentration :

$\tt =\dfrac{3.2~g}{50~cm^3}=0.064~g/cm^3$

For the concentration of Hydrogen chloride in g per dm³ :

$\tt =0.064\dfrac{g}{cm^3}\times \dfrac{1000~cm^3}{dm^3}\\\\=64~g/dm^3$

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

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4 years ago

What does Le Châtelier's principle state?

bonufazy [111]

When a system experiences a disturbance ( such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

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3 years ago

The total number of electrons in the outer shell of a sodium atom is?<br>A. 1<br>B.2<br>C.8<br>D.18

mrs_skeptik [129]

Answer:

Explanation:

outershell atoms of an element are also known as valency of that element

so the valency and number of elctron in the outershell of a sodium atom is +1.

hope this will help

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3 years ago

Which statement describes the blood type of a person with the alleles Ai?

lyudmila [28]

Answer:

d i dont now but im sure my answer is d im not good to write english word

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3 years ago

The element lanthanum has two stable isotopes, lanthanum-138 with an atomic mass of 137.9 AMU and lanthanum-139 with an atomic m

dmitriy555 [2]

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.

Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.

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:

<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>

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3 years ago