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Solnce55 [7]
3 years ago
14

What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275 K? For m, use 0.0399 kg. (1 point)

Chemistry
1 answer:
valentinak56 [21]3 years ago
6 0

What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275 k for m, use 0.0399kg

Answer: The average velocity of the atoms 847.33 m/s.

Explanation:

Moles of the neon = 1.00

Temperature of the gas : 288 K

Mass of the gas = 0.01000

R = 8.31 J/mol K

The average velocity of the atoms 847.33 m/s.

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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
Fed [463]

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

7 0
3 years ago
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4,
Sav [38]

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

  • 2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We <u>convert mass of phthalate to moles</u>, using its molar mass:

  • 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:

  • 9.27 mmol HA * \frac{1mmolBa(OH)_{2}}{2mmolHA} = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

  • 6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So<u> the moles of HCl that reacted </u>are:

  • 17.1 mL * 0.129 M * \frac{2mmolHCl}{1mmolBa(OH)_2} = 4.41 mmol HCl

And the <u>molarity of the HCl solution is</u>:

  • 4.41 mmol / 18.6 mL = 0.237 M

3 0
3 years ago
Jimmy mixes 2 chemicals in the lab. The chemicals change color and they get hot. The rise in temperature
gladu [14]

Answer:

A chemical reaction.

Explanation:

A change in temperature is evidence of a chemical reaction.

Also: They are chemicals...

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3 years ago
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5 0
3 years ago
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