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Serggg [28]
3 years ago
15

2 points

Chemistry
1 answer:
quester [9]3 years ago
6 0

(b) 12.7 L

<u>Explanation:</u>

Given:

Volume, V₁ = 12 L

Pressure, P₁ = 760 mm Hg (at STP)

Temperature, T₁ = 273 K (at STP)

Pressure, P₂ = 720 mm Hg

Temperature, T₂ = 0°C = 273 K

Volume, V₂ = ?

We know:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Substituting the value we get:

\frac{760 X 12}{273} = \frac{720 X V_2}{273} \\\\V_2 = \frac{760 X 12}{720} \\\\V_2 = 12.7 L

Therefore, V₂ is 12.7 L

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Yes they are what are your options
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The Kp for the reaction below is 1.49 × 108 at 100.0°C:
Anna007 [38]
Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8

1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))

COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm
5 0
3 years ago
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

6 0
3 years ago
A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts
Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

                    1 quarts = 0.95 liter

                   7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

        Given:

            1 liter of solution contains 61.2g of barium nitrate:

          7.125 liter will contain  7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

  In solution we have  Ba²⁺ + NO₃⁻

                Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

 

Number of moles of Ba(NO₃)₂ = \frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = \frac{436.05}{271} = 1.609mole

      1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

    therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions  x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0
3 years ago
Science help me pleasee :)
vladimir2022 [97]

Explanation:

during this reaction this will produce 2 molecules of No

8 0
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