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ozzi
3 years ago
7

A 15.0-L vessel contains 0.50 mol CH4 with a pressure of 1.0 atm. After 0.50 mol C2H6 is added to the vessel, what is the partia

l pressure of CH4? The temperature remains unchanged throughout the process.
Chemistry
2 answers:
Kazeer [188]3 years ago
7 0

The partial pressure of methane in the mixture of methane and ethane has been 1 atm.

Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.

The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.

The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.

However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.

For more information about the partial pressure, refer to the link:

brainly.com/question/14623719

allsm [11]3 years ago
5 0

Answer: Hi! The partial pressure of CH4 is

=1 atm

Explanation:

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Bethany has a fish tank at her house. She uses the following data table to keep track of the nitrate and phosphate levels in the
malfutka [58]
<span>sudden increase of Nitrate and Phosphate in March to April that occurred, it causes the fish tank to produce algae</span>
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3 years ago
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(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 10g Na + 10g Oxygen.
Lilit [14]

13.5g

Explanation:

Given parameters:

Mass of Na = 10g

Mass of O₂ = 10g

Unknown:

Mass of products formed = ?

Balanced equation = ?

Solution:

The balanced chemical equation is shown below:

                  4Na       +     O₂     ⇒      2Na₂O

In any reaction, the specie in short supply determines the extent of the reaction.

This reaction is not an exclusion. We need to first determine the specie in short supply and use it to estimate the amount of product since we have a 100% yield which signifies that all was used up.

  let us convert to moles;

    Number of moles of Na = \frac{mass }{molar mass}  = \frac{10}{23} = 0.435mole

 Number of moles of O₂ = \frac{mass}{molar mass} = \frac{10}{32} = 0.313mole

From the given equation;

   4 moles of Na requires 1 mole of O₂;

  0.435 moles of Na will require \frac{0.435}{4} = 0.11 moles

 But the given amount O₂ is 0.313, this is an excess of 0.313 - 0.11 = 0.203moles

We see that Na is the limiting reagent;

   4 moles of Na gives 2 mole of Na₂O

   0.435 moles of Na will give \frac{0.435 x 2 }{4} = 0.22 moles

Mass of Na₂O = number of moles x molar mass = 62 x 0.22 = 13.5g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

5 0
4 years ago
In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is
GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

ΔG = RT In ( a2 / a1 )  

4 0
3 years ago
How many moles of aluminum oxide al2o3 are in a sample with a mass of 204.0
erik [133]

Answer:

2 moles

Explanation:

Let us first start by calculating the molecular mass of Al₂O₃.

The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.

Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

Therefore, molecular mass of Al₂O₃ is:

= (2*26.981539) + (3*15.999) u

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

3 0
3 years ago
Read 2 more answers
Help me in my this plzzz ​
eduard

Answer:

a is oxidation

b is reduction

c is reduction

d is oxidation

hope it helps you

8 0
3 years ago
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