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Klio2033 [76]
3 years ago
10

The volume of a gas is 550 mL at 960 mm Hg and 200.0 C. What volume

Chemistry
1 answer:
Masja [62]3 years ago
4 0

Answer:

The volume will be 568.89 mL.

Explanation:

Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Where P = pressure, T = temperature, K = Constant

Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Studying an initial state 1 and a final state 2, it is fulfilled:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 960 mmHg
  • V1= 550 mL
  • T1= 200 C= 473 K (being 0 C=273 K)
  • P2= 830 mmHg
  • V2= ?
  • T2= 150 C= 423 K

Replacing:

\frac{960 mmHg*550 mL}{473K} =\frac{830 mmHg*V2}{423 K}

Solving:

V2=\frac{423 K}{830 mmHg} *\frac{960 mmHg*550 mL}{473K}

V2= 568.9 mL

<u><em>The volume will be 568.89 mL.</em></u>

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2. How many grams of NaCl are required to prepare 0.40 L of a 0.75 M solution?
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The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles of a solution must be initially calculated by using the following formula:

molarity = no of moles ÷ volume

no of moles = 0.75 × 0.40

no of moles = 0.3 moles

mass of NaCl = 0.3 × 58.5 = 17.55g

Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.

Learn more about mass at: brainly.com/question/19694949

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
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Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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