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3 years ago

Anthony's homework assignment is to demonstrate that an orange has already undergone a chemical change. Which of the following

Chemistry

2 answers:

wolverine [178]3 years ago

8 0

B. Rotten orange is the correct answer. Hope this helps!

Wittaler [7]3 years ago

5 0

I agree, it’s B because you can’t change a rotten orange back into a fresh orange.

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A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partia

viva [34]

Explanation:

The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.

Total pressure = 2atm

Mole Fraction = number of moles / total number of moles

Neon

Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

Argon

Mole Fraction = 0.74 / 7.35 = 0.101

Partial Pressure = 0.101 * 2 = 0.202 atm

Xenon

Mole Fraction = 2.15 / 7.35 = 0.293

Partial Pressure = 0.293 * 2 = 0.586 atm

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3 years ago

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What are oil and waters properties of matter?

yulyashka [42]

Answer:

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3 years ago

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

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