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Yuliya22 [10]
3 years ago
10

A gaseous hydrocarbon contains 80% carbon and 20% hydrogen, 1dm3 of the compound at s.t.p weighs 1.35g find the molecular formul

ar (1 mole of any gas at s.t.p occupies 22.4dm3, C=12, H=1)​
Chemistry
1 answer:
Cloud [144]3 years ago
4 0

Answer:

C₂H₆

Explanation:

To obtain molecular formula of the gas, you need to find first its molecular mass.

To find molecular mass you need to obtain moles of gasbecause molecular mass is the ratio between mass of the gas and the volume it occupies.

As at STP, 1 mole of a gas occupies 22.4dm³, 1dm³ of gas are:

1dm³ × (1mol / 22.4dm³) = 004464moles.

Its molecular mass is:

1.35g / 004464moles =

30.24g/mol

Now, if the gas is 80%C and 20%H, its <em>empirical formula (Simplest ratio of atoms in a molecule) </em>is:

80% Carbon × (1mol / 12.01g) = 6.66moles C

20% Hydrogen × (1mol / 1.01g) = 19.8moles H

Ratio of H:C is:

19.8 mol H / 6.66mol C = 3

Thus, you can know you will have 3 moles of Hydrogen per mole of Carbon, CH₃ (It weighs 15.04g/mol)

As the gas weighs 30.24 ≈ 2 CH₃

The molecular formula is:

<h3>C₂H₆</h3>
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What are some limitations of this experiment? How could it be improved?
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A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2
mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: 2NO \Leftrightarrow N_{2}O_{2}    (fast)

Rate expression for step 1 is as follows.

               Rate = k [NO]^{2}

Step 2: N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O    (fast)

Here, N_{2}O_{2} is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0
3 years ago
a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses
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Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

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P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa

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2 years ago
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