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kari74 [83]
2 years ago
10

A balloon filled with helium has a volume of 30.0 L at a pressure of 100 kPa and a temperature at 15.0 Celsius. What will The vo

lume of the balloon be if the temperature is increased at 80.0 Celsius and the pressure changes to 115 kPa?
Chemistry
2 answers:
Anettt [7]2 years ago
3 0

Answer:

The answer to your question is V2 = 31.97 L

Explanation:

Data

Volume 1 = 30 L

Pressure 1 = 100 kPa

Temperature 1 = 15°C

Volume 2 = ?

Pressure 2 = 115 kPa

Temperature 2 = 80°C

Process

1.- Convert temperature to °K

Temperature 1 = 15 + 273 = 288°K

Temperature 2 = 80 + 273 = 353°K

2.- Use the Combined gas law to solve this problem

                  P1V1/T1 = P2V2/T2

-Solve for V2

                 V2 = P1V1T2 / T1P2

-Substitution

                 V2 = (100 x 30 x 353) / (288 x 115)

- Simplification

                 V2 = 1059000 / 33120

- Result

                 V2 = 31.97 L

defon2 years ago
3 0

Answer:

The new volume is 31.97 L

Explanation:

Step 1: Data given

The initial volume = 30.0 L

The initial pressure = 100 kPa

Temperature = 15.0 °C = 288 K

The temperature is increased to 80.0 °C = 353 K

The pressure increases to 115 kPa

Step 2: Calculate the new volume

P1*V1 /T1  = P2*V2/T2

⇒with P1 = the initial pressure = 100 kPa

⇒with V1 = the initial volume = 30.0 L

⇒with T1 = the initial temperature = 288 K

⇒with P2 = the increased pressure = 115 kPa

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the increased temperature = 353 K

(100 * 30.0)/288 = (115 * V2)/353

10.417 = (115 * V2)/353

V2 = 31.97 L

The new volume is 31.97 L

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Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

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5 0
2 years ago
At 50 degree C pK_w = 13.26. What is the pH of pure water at this temperature?
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Answer:

6.63

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From the relationship;

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Pkw = 13.26

Then it follows that;

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Hence if. pKw = 13.26, the pH = 6.63

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A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
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Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
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When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

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