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BartSMP [9]
3 years ago
15

Physical properties for helium?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
4 0
Helium is a colorless, odorless, tasteless gas. It has a number of unusual properties. For example, it has the lowest boiling point of any element, -268.9°C (-452.0°F). The boiling point for a gas is the temperature at which the gas changes to a liquid.
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How many moles of CF4 can be produced when 8.95 mol C reacts with 7.88 mol F2?
Molodets [167]

The remaining moles of C is 5.01 moles while the remaining moles of F₂ is 0.

<h3>Reaction between Carbon and Fluorine </h3>

The reaction between carbon and Fluorine is given as;

C + 2F₂  -------> CF₄

1  :    2                1

from the reaction above,

2 moles of F₂ requires 1 mole of C

7.88 mole of  F₂ will require:    7.88/2 = 3.94 moles of C and 3.94 moles of CF₄.

The remaining moles of C = 8.95 - 3.94 = 5.01 moles while the remaining moles of F₂ is 0.

Learn more about moles here: brainly.com/question/15356425

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divide the distance by the average

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3 years ago
Which of the following is a mixture
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Gold Ore is a mixture.
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3 years ago
Read 2 more answers
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.

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If someone adds million of small fish to a lake, how would the number of big fish change
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The population of big fish would increase because they have more food
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