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pogonyaev
3 years ago
5

Upon decomposition, one sample of magnesium fluoride produced 1.65 kg of magnesium and 2.56 kg of fluorine. A second sample prod

uced 1.32 kg of magnesium. Part A How much fluorine (in grams) did the second sample produce?
Chemistry
1 answer:
aleksley [76]3 years ago
7 0

Answer : The mass of fluorine in second sample produces, 2.048 grams

Law of definite proportion states that in a chemical compound, the components of the element are in fixed ratio.

The given chemical compound is magnesium fluoride. In the magnesium fluoride the components of the element magnesium and fluorine are in fixed ratio that is, 1 : 1

\frac{M_{Mg}}{M_{F}}=\frac{m_{Mg}}{_{F}}

M_{Mg} = mass of magnesium in sample 1 = 1.65 g

M_{F} = mass of fluorine in sample 1 = 2.56 g

m_{Mg} = mass of magnesium in sample 2 = 1.32 g

m_{F} = mass of fluorine in sample 2 = x

Now put all the give values in the above relation, we get

\frac{1.65g}{2.56g}=\frac{1.32g}{x}

x=2.048g

Therefore, the mass of fluorine in second sample produces, 2.048 grams

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Furkat [3]

Answer:

The resulting pressure in the flask is 0.93 atm

Explanation:

- Apply the Ideal Gas law in both cases to get the mols, of Ar and SO2.

- Once you know the mols, sum both to get the total mols in the mixture.

- Apply the Ideal Gas lawin the flask with the total mols to know the resulting pressure.

First: 0.750 L of argon at 1.50 atm and 177°C

T° C + 273 = T° K → 177°C + 273 = 450K

P .V = n . R . T

1.50 atm . 0.750 L = n . 0.082 L.atm/mol.K  . 450K

(1.50 atm . 0.750 L) /  (0.082 mol.K/L.atm  . 450K) = n

0.030 mols Ar = n

Be careful with the R units, the ideal gases constant

Let's convert kPa to atm.

101.33 kPa _____ 1 atm

95 kPa ________ (95 / 101.33) = 0.94 atm

T° C + 273 = T° K → 63°C + 273 = 336 K

0.94 atm . 0.235 L = n . 0.082 L.atm/mol.K . 336K

(0.94 atm . 0.235 L) / (0.082 mol.K/L.atm . 336K) = n

8.01X10⁻³ mols = n

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1L . P = 0.038 mol . 0.082 L.atm / mol.K . 298 K

P = (0.038 mol . 0.082 L.atm / mol.K . 298 K ) / 1L

P = 0.93 atm

7 0
2 years ago
A+gas+at+250k+and+15atm+has+a+molar+volume+12%+less+than+the+perfect+gas+law+character. +calculate+the+compressibility+factor
shutvik [7]

The compression factor based on the information given is 0.083.

<h3>How to calculate the compression factor?</h3>

The way to calculate the compression factor goes this:

V(ideal) = RT/P

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V(real) will be:

= V(ideal)/12

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The compression factor will be:

= 1.368/(12 × 1.368)

= 0.083

Therefore, the compression factor is 0.083.

Learn more about compression factor on:

brainly.com/question/24261456

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If the temperature of the reaction mixture is decreased,then the equilibrium will shift to increase the temperature.

During the formation of the ammonia,it gives off heat.So it is an exothermic reaction.

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A decrease in the temperature favors the reaction that is exothermic (the forward reaction)because it produces energy.Therefore,if the temperature is decreased,the yield of the ammonia increases.

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