A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula:

Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n

lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

$\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239$

According to Dalton's Law of partial pressure-

$P_i=\chi_i\times P_{total}$

Where,

$P_i=$ The pressure of the gas component in the mixture

$\chi_i=$ Mole fraction of that gas component

$P_t=$ The total pressure of the mixture

$P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm$

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

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

$n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole$

Learn more about Dalton's Law of partial pressure here;

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4 0

1 year ago

A biotic factor in a desert ecosystem would be the

valkas [14]

C is correct.

Please vote my answer brainliest. thanks!

6 0

3 years ago

Read 2 more answers

How do the percent compositions for C3H6 and C4H7 compare?

mariarad [96]

A. They are the same

<h3>Further explanation</h3>

Given

C3H6 and C4H8

Required

The percent compositions

Solution

C₃H₆(MW = 42 g/mol)

%C = 3.12/42 x 100% = 85.71%

%H = 6.1/42 x 1005 = 14.29%

C₄H₈(MW=56 g/mol)

%C = 4.12/56 x 100% = 85.71%

%H = 8.1/56 x 100%=14.29%

So they are the same, because mol ratio of C and H in both compounds is the same, 1: 2

3 0

3 years ago

what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?

nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0

3 years ago

A mixture of CuSO4 · 5H2O and MgSO4 · 7H2O is heated until all the water is lost. If 5.127 g of the mixture gives 2.817 g of the

Nadusha1986 [10]

Answer:

Let the mixture is X% by mass of CuSO

4

.5H

2

O and 100 - X % by mass of MgSO

4

.7H

2

O. 5.0 g of mixture will contain 0.05X g CuSO

4

.5H

2

O and 5.0 - 0.05X g MgSO

4

.7H

2

O

The molar masses of CuSO

4

.5H

2

O and MgSO

4

.7H

2

O are 249.7 g/mol and 246.5 g/mol respectively.

The number of moles of CuSO

4

.5H

2

O=

249.7

0.05X

=2.00×10

−4

X moles.

Explanation:

Pls mark it as branliest answere thanks

3 0

3 years ago