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2 years ago

How many grams of Cl are in 38.0 g of each sample of chlorofluorocarbons (CFCs)?

1 answer:

5 0

If there are 38 g of CFC, then there will be 120.9135 g of CFCl2 per mol. You will multiply this times the number of moles of Cl (2) for every mole of CFCl2, and then by the number of grams of Cl per mole, which is 35.4532: (38.0 g CF2Cl2) / (120.9135 g CF2Cl2/mol) x (2 mol Cl / 1 mol CF2Cl2) x (35.4532 g Cl/mol) = 22.3 g Cl in CF2Cl2 Next, if there are 38 g of CFC, there will be 137.3861 g of CFCl3 per mole. You will multiply this times the number of moles of Cl (3 this time) for every mole of CFCl3. You will then multiply this by 35.4532 again: (38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) = 29.4 g Cl in CFCl3 Continue following these steps until you are able to multiply 1 mole of Cl per 1 mol CF3Cl by 35.4532: (38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) = 21.6 g Cl in C2F3Cl3 (38.0 g CF3Cl) / (104.4589 g CF3Cl/mol) x (1 mol Cl / 1 mol CF3Cl) x (35.4532 g Cl/mol) = 12.9 g Cl in CF3Cl

You might be interested in

What is the name of the functional group that is attached to this hydrocarbon? The first and last of a chain of three carbons ar

disa [49]

Answer:

Ketone

Explanation:

As you are stating here, we have a carbonated chain of three carbons, and the first and last has 3 Hydrogens, then this means that we have CH₃ . The center carbon is a carbon double bonded to oxygen.

In general terms this belongs to the carbonyl group. However, this alone does not represent a functional group, but when it's in a chain with other radycals or chains, it becomes a functional group.

In this case, the molecule you are talking here is the following:

CH₃ - CO - CH₃

This molecule is known as the Acetone, and has the general form of:

R - CO - R'

Which belongs to a ketone as a functional group.

4 0

1 year ago

Why does 4.03/0.0000035 = 1.2 x 106, instead of a different number of significant figures?

jasenka [17]

Explanations:- As per the significant figures rule, In multiplication and division, we go with least number of sig figs.

4.03 has three sig figs where as 0.0000035 has two sig figs only, The zeros in this number are not sig figs as they are just holding the place values. As the least number of sig figs here is two, the answer needs to be reported with two sig figs only.

$\frac{4.03}{0.0000035}=1.2*10^6$

4 0

2 years ago

the image above shows a chamber with a fixed volume filled with gas at a pressure of 1560 mmHg and a temperature of 445.0 K. If

Sedaia [141]

Answer:

The new pressure of the gas in the chamber is 1,093.75 mmHg

Explanation:

The Gay-Lussac Law is a gas law that relates pressure and temperature to constant volume. This law says that the pressure of the gas is directly proportional to its temperature.

That is, if the temperature increases, the pressure increases, while if the temperature decreases, the pressure decreases. So the Gay-Lussac law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Having an initial and an end state of a gas, the following expression can be used:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 1560 mmHg
T1= 445 K
P2=?
T2= 312 K

Replacing:

$\frac{1560 mmHg}{445 K} =\frac{P2}{312 K}$

Solving:

$P2=\frac{1560 mmHg}{445 K} *312K$

P2=1,093.75 mmHg

The new pressure of the gas in the chamber is 1,093.75 mmHg

7 0

1 year ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

3 0

2 years ago

When the reaction begins, the researcher records that the rate of reaction is such that 1 mole of A is consumed per minute. Afte

yanalaym [24]

Answer:

C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy.

Explanation:

For the reaction:

A + B → 2C

You can increase the consume of A:

A. Utilize reaction conditions to convert both reactants to solids. FALSE. Convert both reactants in solids makes less frequent collisions doing difficult the reaction.

B. Decrease the concentration of reactant B to allow C to be produced at a greater rate. FALSE. The decreasing of B will make the reaction slowly.

C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy. TRUE. The increase of temperature will make more frequent collisions of A and B doing the reaction faster.

D. Increase the concentration of C to allow for more frequent collisions of A and B of higher energy . FALSE. The increase of C makes less frequent collisions doing difficult the reaction.

E. Introduce a catalyst to decrease the consumption of A and B. FALSE. Introducing a catalyst increase the consumption of A and B.

I hope it helps!

8 0

1 year ago