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2 years ago

In a 66.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.290. What is the mass of each component? g

Chemistry

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

23.84g CH30H

32.81g H2O

Explanation:

We will be using the definition of mole fraction to determine the relationship between the number of moles of methanol,

CH3OH , and the number of moles of water.

But mole fraction gives the ratio between the number of moles of a component i of a solution to the total number of moles present in that solution.

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

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Which of the following best describes the changes involved in a fission reaction?

never [62]

The correct answer is this: THE NUCLEUS OF AN ATOM SPLITS INTO FRAGMENTS, RELEASING A LARGE AMOUNT OF ENERGY.
Nuclear fission is the process in which the nucleus of a radioactive element split into two different nucleic of smaller sizes of different elements with a large release of energy. Nuclear fission process is usually used to provide energy for electricity generation.

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2 years ago

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

2 years ago

Can somebody plz help answer these questions plz (grwde7science) only if u know how to do it thx :3

dem82 [27]

1. Crystallization. Magma cools either underground or on the surface and hardens into an igneous rock.
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4. Rock Melting.
Metamorphic rocks underground melt to become magma. If you need more help https://www.ck12.org/earth-science/rock-cycle-processes/lesson/Rocks-and-Processes-of-the-Rock-Cycle-HS-ES/

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2 years ago

You may have heard the saying, “You are what you eat.” Use information learned in this section to explain what this statement me

Ilia_Sergeevich [38]

You got to be greatfull for what you get to eat .

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2 years ago

Can someone help with this please? My teacher isnt responding to any of mine or my classmates requests for help

sergiy2304 [10]

Answer:

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2 years ago

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