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Jobisdone [24]
2 years ago
8

A balloon has a pressure of 3.1 atm at a volume of 155 ml. if the temperature is held constant, what is the volume (ml) if the p

ressure is increased to 10.5 atm?
Chemistry
1 answer:
GarryVolchara [31]2 years ago
4 0

Answer:

45.8 mL

Explanation:

If all variables are held constant, the new volume can be found using the Boyle's Law equation. The equation looks like this:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new volume by plugging the given values into the equation and simplifying.

P₁ = 3.1 atm                        P₂ = 10.5 atm

V₁ = 155 mL                       V₂ = ? mL

P₁V₁ = P₂V₂                                                 <----- Boyle's Law equation

(3.1 atm)(155 mL) = (10.5 atm)V₂                <----- Insert values

480.5 = (10.5 atm)V₂                                  <----- Multiply 3.1 and 155

45.8 = V₂                                                    <----- Divide both sides by 10.5

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4 0
2 years ago
Using 18.8L, calculate the volume (in L) of water vapour that should be produced by the reaction of 35.0 g
victus00 [196]

Answer:

A. 18.8L

B. 75.2L of H2O.

Explanation:

A. Determination of the volume of 35g of C3H8.

Date obtained from the question include the following:

Mass of C3H8 = 35g

Temperature (T) = 40°C

Pressure (P) = 110KPa

Volume (V) =..?

Next, we shall determine the number of mole (n) in 35g of C3H8. This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mass of C3H8 = 35g

Mole of C3H8 =..?

Mole = mass /molar mass

Mole of C3H8 = 35/44

Mole of C3H8 = 0.795 mole

Finally, we shall determine the volume of 35g of C3H8 as follow:

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 110KPa

Number of mole (n) = 0.795 mole

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =..?

PV = nRT

110 x V = 0.795 x 8.314 x 313

Divide both side by 110

V = (0.795 x 8.314 x 313)/110

V = 18.8L

Therefore, the volume of 35g of C3H8 under the conditions given is 18.8L

B. Determination of the volume of water vapour produced by the reaction of 35g of propane, C3H8.

From the calculations made in (A) above, 35g of C3H8 is equivalent to 18.8L of C3H8.

Thus, we can obtain the volume of water vapour produced as follow:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the balanced equation above,

1L of C3H8 reacted to produce 4L of H2O.

Therefore, 18.8L of C3H8 will react to produce = (18.8 x 4)/1 = 75.2L of H2O.

Therefore, 75.2L of H2O were produced from the reaction.

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3 years ago
Which of the following steps in a star's life cycle takes the longest?
Vsevolod [243]
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3 years ago
How do you prepare 1 liter of a 0.5 molar glucose solution?
riadik2000 [5.3K]
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0.5 molar of glucose means 0.5mole glucose for 1 liter of water. Since we want to made 1L of solution, then the amount of glucose needed is: 0.5mole/l x 1l= = 0.5mole

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6 0
3 years ago
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svlad2 [7]

Answer:

The resulting solution is basic.

Explanation:

The reaction that takes place is:

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First we <u>calculate the added moles of HNO₃ and KOH</u>:

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As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.

8 0
2 years ago
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