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3 years ago

How many moles are equal to 83.4 L of O2?

Chemistry

1 answer:

ladessa [460]3 years ago

8 0

Answer:

$3.72mol$

Explanation:

Hello,

In this case, we consider that at STP conditions (273 K and 1 atm) we know that the volume of 1 mole of a gas is 22.4 L, thereby, for 83.4 L, the resulting moles are:

$22.4L\rightarrow 1mol\\83.4L\rightarrow X\\X=\frac{83.4L*1mol}{22.4L}=3.72mol$

This is a case in which we apply the Avogadro's law which relates the volume and the moles as a directly proportional relationship.

Best regards.

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B. The equilibrium constant for the reaction 2H2(g) + S2(g) 2H2S(g) is Keq = . (3 points)

eimsori [14]

Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

Hello,

i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

$Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99$

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

Best regards.

4 0

2 years ago

Which of the following would be more likely to conduct electrons? Strontium or Iodine? Defend your choice.

Alchen [17]

Strontium because it is a metal

6 0

2 years ago

What specific electrophile does benzene attack when it undergoes nitration??

Deffense [45]

The reaction is as follows:
C6H6 + HNO3 + H2SO4 ------------> C6H5NO2 + H2O
(BENZENE) (NITRIC ACID)(CATALYST)
NO2(+) is the electrophile that acctacks on the benzene ring in nitration process.

7 0

3 years ago

When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?

stepladder [879]

Answer:

$m_{precipitated}=24.8g$

Explanation:

Hello,

In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

$m_{precipitated}=108g-83.2g\\\\m_{precipitated}=24.8g$

Best regards.

8 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

2 years ago