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blsea [12.9K]
2 years ago
6

A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m

oles of CO2 would be formed if 3moles of the compound react with oxygen?
Chemistry
1 answer:
Phoenix [80]2 years ago
7 0
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
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A generic gas, x, is placed in a sealed glass jar and decomposes to form gaseous y and solid z. 2x(g)↽−−⇀y(g)+z(s) how are these
Aleksandr [31]
Answer: the equilibrium will be displaced to the right leading an increase on the quantities of y(g) and z(s).

Justification:

According to the rules of equilibrium, based on Le Chatellier's priciple, any change in a system in equilibrium will be tried to be compensated to restablish the equilibrium

The higher the amount, and so the concentration, of X(g), the more will the forward reaction proceed leading to an increase on the concentration of the products y(g) and z (s). Look that that will also be accompanied by a decreasing on the pressure, since 2 molecules of the gas X(g) are converted into 1 molecule of the gas y(g).
3 0
2 years ago
Read 2 more answers
in a candy factory, the nutty chocolate bars contain 22.0% by mass pecans. If 5.0 kg of pecans were used for candy last Tuesday,
zysi [14]
22.0 is the same as saying that in 100 grams of a chocolate bar, there are 22.0 grams of pecans. or to make it easier because of this problem- 100 Kilograms of a chocolate bar, there is 22.0 Kg of pecans. we can use this as a conversion factor (what is used to convert a value to another value. 

conversion factor---> 22.0 kg of pecan= 100 kg of chocolate bar
Note: remember this, what you are converting from goes in the denominator, what you converting to goes in the numerator. 

5.0 Kg of pecan (100 Kg of chocolate bar/ 22.0 Kg of pecan)= 23 Kg of chocolate bar


7 0
2 years ago
A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with
andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g

Molar mass of nitrogen gas, N_2 = 28.014 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass}{28.014\ g/mol}

Mass of nitrogen gas = 28.014x g

So,

Let, Mass_{Container}=y

y+28.014x=37.289

Similarly,

Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g

Molar mass of nitrogen gas, CO_2 = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass{44.01\ g/mol}

Mass of nitrogen gas = 44.01x g

So,

y+44.01x=37.440

Solving the two equations, we get :

Mass_{Container}=y=37.025\ g

x = 0.00943 moles

Thus, Given:

Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g

37.025\ g+Mass_{Unknown\ gas}=37.062\ g

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00943\ moles= \frac{0.037\ g}Molar mass}

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0
2 years ago
Groups of atoms that are added to carbon backbones and give them unique properties are known as
Irina-Kira [14]

Answer:

             Groups of atoms that are added to carbon backbones and give them unique properties are known as <u>Functional Groups</u>.

Explanation:

                   In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.

Examples:

S No.                          Functional Group                                 Name

1                                   R--X                                             Alkyl Halides

2                                   R--OH                                          Alcohols

3                                  R--NH₂                                         Amines

4                                  R--O--R                                         Ethers

5                                   R--CO--R                                      Ketones

6                                   R--CO--H                                     Aldehydes

7                                  R--CO--OH                                  Carboxylic acids

8                                   R--CO--X                                     Acid Halides

10                                R--CO--NR₂                                 Acid Amides

11                                 R--CO-OR'                                  Esters

3 0
2 years ago
In a group assignment, students are required to fill 10 beakers with 0.720 M CaCl2. If the molar mass of CaCl2 is 110.98 g/mol a
8_murik_8 [283]
The answer is 200 g.

If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M 
x = 79.90 g

So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be  79.90 g.

Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L

79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
8 0
2 years ago
Read 2 more answers
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