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agasfer [191]
3 years ago
14

A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 113.0 L and a pressur

e of 122.0 atm at 25.0 °C. After a while, she notices that the valve has not been closed properly. The pressure had dropped to 112.0 atm (The tank is still at 25.0 °C).
Chemistry
1 answer:
lisov135 [29]3 years ago
7 0
Given Data:
                   P₁  =  122 atm

                   P₂  =  112 atm

                   V₁  =  113 L

                   V₂  = ?

Solution:
              Let suppose the gas is acting ideally. According to Ideal gas equation, keeping the temperature constant,

                                  P₁ V₁  =  P₂ V₂
Solving for V₂,
                                      V₂  =  P₁ V₁ / P₂
Putting Values,
                                      V₂  =  (122 atm × 113 L) ÷ 112 atm
                 
                                      V₂  =  113 L

Converting Volume to Moles,
 As,
                         22.4 L is occupied by  =  1 mole of He gas
So,
                         113 L will occupy        =  X moles of He gas

Solving for X,
                                     X  =  (113 L × 1 mole) ÷ 22.4 L

                                     X  =  5.04 Moles of He 
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Calculations
Black_prince [1.1K]

Answer:

Theoretical yield: 2.75g of paranitroaniline

Percentage yield: 54.5%

Explanation:

In strong-acid medium, acetanilide (Molar mass: 135.16g/mol) reacts producing para-nitroaniline (138.12g/mol) in a 1/1 reaction.

Theoretical yield of para-nitroaniline is the mass produced assuming a yield of 100%. That is:

2.7g acetanilide × (1mol / 135.16g) = 0.020 moles of acetanilide.

Assuming a yield of 100% are 0.020 moles of paranitroaniline. Theoretical yield is:

0.020 moles × (138.12g / mol) =

<h3>Theoretical yield: 2.75g of paranitroaniline</h3><h3 />

Percentage yield is:

(Actual yield / theoretical yield) × 100

Actual yield was 1.5g and percentage yield will be:

Percentage yield: (1.5g / 2.75g) × 100

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8 0
3 years ago
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A 3.3 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.3 g. The two substances react, rel
Zanzabum

Answer:

1.73g of CO2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass of CH3COOH from the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.

Next, we shall determine the limiting reactant of the reaction. This is illustrated below:

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.

From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.

Finally, can determine the mass of CO2 produced during the reaction.

In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:

From the balanced equation above,

84g of NaHCO3 reacted to produce 44g of CO2.

Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.

Therefore, 1.73g of CO2 is released during the reaction.

7 0
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