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3 years ago

What is the molar mass of an unknown hydrocarbon whose density is measured to be 1.97 g/l at stp?

Chemistry

2 answers:

notka56 [123]3 years ago

8 0

Answer:

The correct answer is

M = 44.1 g / mol

Explanation:

Hello!

Let's solve this!

First we write down the data

d: 1.97 g / L

P: 1 atm

R: 0.082 atmL / Kmol

T: 273K

From the ideal gas formula:

d = P * M / R * T

We cleared M to know the molar mass.

M = (d * R * T) / P

M = (1.97 g / L * 0.082 atmL / Kmol * 273K) / 1atm

The correct answer is

M = 44.1 g / mol

inessss [21]3 years ago

5 0

Answer: The correct answer is : 44.1 g / mol

Explanation: d = PM / RT

M = dRT / P

d = 1.97 g / l

R = ideal gas (0.082 06 l - atm / mol - ° K)

P = 1 atm

T = 273 ° K

M = Molar mass

M = 1.97 g / l x 0.082 l - atm / mol - ° K x 273 ° K / 1 atm

M = 44.1 g / mol

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2 years ago

Read 2 more answers

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

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5 0

2 years ago

A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa a

Damm [24]

Answer:

71.92 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

T1 = 50°C = 50 + 273 = 323K

V1 = 105L

T2 = -25°C = -25 + 273 = 248K

P2 = 105.4 kPa

P1 = ?

V2 = 55.0 L

Using P1V1/T1 = P2V2/T2

P1 × 105/323 = 105.4 × 55/248

105P1/323 = 5797/248

0.325P1 = 23.375

P1 = 23.375 ÷ 0.325

P1 = 71.92 kPa

4 0

2 years ago