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3 years ago

10

It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it

take him to jog to Chicago if he jogged continuously and REALLY wanted to get there?

1 answer:

Lina20 [59]3 years ago

5 0

Answer:

3,048 minutes

Explanation:

762 divided by 5

that number times 20

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A rigid tank that contains 2 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25°C an

koban [17]

Answer:

The volume in the first tank = 0.32 $m^{3}$

The volume in the second tank = 2.066 $m^{3}$

The final pressure of the mixture = 203.64 K pa

Explanation:

<u>First Tank </u>

Mass = 2 kg

Pressure = 550 k pa

Temperature = 25 °c = 298 K

Gas constant for nitrogen = 0.297 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

550 × V = 2 × 0.297 × 298

V = 0.32 $m^{3}$

This is the volume in the first tank.

<u>Second tank</u>

Mass = 4 kg

Pressure = 150 K pa

Temperature = 25 °c = 298 K

Gas constant for oxygen = 0.26 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

150 × V = 4 × 0.26 × 298

V = 2.066 $m^{3}$

This is the volume in the second tank.

This is the iso thermal mixing. i.e.

$P_{3} V_{3} = P_{1} V_{1} + P_{2} V_{2}$ ----- (1)

$V_{3} = V_{1} + V_{2}$

$V_{3} = 0.32 + 2.066$

$V_{3} = 2.386 \ m^{3}$

Put this value in equation (1)

$P_{3}$ × 2.386 = 550 × 0.32 + 150 × 2.066

$P_{3}$ = 203.64 K pa

Therefore the final pressure of the mixture = 203.64 K pa

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what is the mole fraction of neon in a mixture that contains 0.628 g of helium, 11.491 g of neon, and 7.613 g of argon?

VMariaS [17]

Answer:

0.2

Explanation:

Given parameters:

Mass of helium = 0.628g

Mass of neon = 11.491g

Mass of argon = 7.613g

Unknown:

Mole fraction of neon = ?

Solution:

The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.

We need to calculate the number of moles of each element first;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Helium = 4g/mol

Molar mass of Neon = 20g/mol

Molar mass of Argon = 40g/mol

Number of moles of He = $\frac{0.628}{4}$ = 0.16moles

Number of moles of Ne = $\frac{11.491}{20}$ = 0.58moles

Number of moles of Ar = $\frac{7.613}{40}$ = 0.19moles

Total number of moles = 0.16moles + 0.58moles + 0.19moles = 0.93moles

Mole fraction Neon = $\frac{0.19}{0.93}$ = 0.2

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