All categories

4 years ago

Neon effuses 1.26 times as fast an an unknown gas at a particular temperature. What is the molar mass of the unknown gas

Chemistry

1 answer:

Svetach [21]4 years ago

8 0

Answer:

31.75 amu

Explanation:

Using Graham's equation of effusion as depicted below:

V1/V2 = √m2/√m1

Where; v1 = speed of neon gas

V2 = speed of unknown gas

m1 = molar mass of neon gas

m2 = molar mass of unknown gas

According to this question: v1 = 1.26x, m1 = 20amu, v2 = 1x, m2 = ?

Hence,

1.26x/1x = √m2/√20

1.26/1 = √m2/4.472

√m2 = 4.472 × 1.26

√m2 = 5.635

m2 = 5.635²

m2 = 31.75 amu

Therefore, the molar mass of the unknown gas is 31.75 amu.

You might be interested in

What volume of ethanol (density = 0.7893 g/cm3) should be added to 450. ml of water in order to have a solution that freezes at

AysviL [449]

According to this formula :
ΔTf= i Kf m
i is van't Hoff factor= 1
Kf = 1.86
m the molality we need to assume it
m= x moles of C2H5OH / Kg of mass
∴ 15 = 1 * 1.86 * ( x moles of C2H5OH/ 0.45 kg)
∴X = 3.629 moles
mass = no.of moles x molar mass of C2H5OH
= 3.629 X 46 = 167 g
∴the volume = mass / denisty
= 167 / 0.7893 = 211.57≈ 212 mL

4 0

4 years ago

Does this sound good so far

bogdanovich [222]

I think it looks good but some grammars are wrong i suggest trying to put it on grammarly and just check it out its just my suggestion and not doing this for points so you can delete this if it doesnt help

4 0

4 years ago

What is the iupac name of N2O4

galben [10]

Dinitrogen Tetraoxide

Formula: N₂O₄

Molar mass: 92.011 g/mol

6 0

3 years ago

Read 2 more answers

What is the value of n?

Softa [21]

M = n/V

M = 1.51
V = 0.730

1.51 = n/0.730

To solve this, multiply both sides by the V.

1.1023 = n

n = 1.1023 mol

4 0

3 years ago

4. Your mission, if you choose to accept it, is to make 10mmol/L acetate buffer, pH5.0. Beginning with 10mmol/L HAc, what concen

julia-pushkina [17]

Answer:

6,45mmol/L of NaOH you need to add to reach this pH.

Explanation:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,74

Henderson-Hasselbalch equation for acetate buffer is:

5,0 = 4,74 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

Solving:

1,82 = $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$ (1)

As total concentration of acetate buffer is:

10 mM = [CH₃COOH] + [CH₃COO⁻] (2)

Replacing (2) in (1)

[CH₃COOH] = 3,55 mM

And

[CH₃COO⁻] = 6,45 mM

Knowing that:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Having in the first 10mmol/L of CH₃COOH, you need to add 6,45 mmol/L of NaOH. to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃COOH .

I hope it helps!

7 0

3 years ago