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Jobisdone [24]
3 years ago
13

Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlori

te (NaOCl). What will be the pH of this buffer solution after adding 0.341 mol of HCl? The Ka of hypochlorous acid is 2.9 x 10-8 . Assume the change in volume of the buffer solution is negligible
Chemistry
1 answer:
Helga [31]3 years ago
6 0

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:

  • pH = pKa + log\frac{[NaOCl]}{[HOCl]}

We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:

  • pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
  • [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
  • pH = 7.54 + log \frac{4}{4}
  • pH = 7.54

By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:

  • pH = 7.54 + log\frac{[NaOCl-HCl]}{[HOCl+HCl]}
  • pH = 7.54 + log \frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}
  • pH = 6.82
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A sample of nitrogen gas collected at a pressure of 1.03 atm and a temperature of 279 K is found to occupy a volume of 568 milli
DIA [1.3K]

Answer: 0.025 moles of nitrogen gas are there in the sample.

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1.03 atm

V = Volume of gas = 568 ml = 0.568 L   (1L=1000ml)

n = number of moles  = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =279K

n=\frac{PV}{RT}

n=\frac{1.03atm\times 0.568L}{0.0821L atm/K mol\times 279K}=0.025moles

0.025 moles of nitrogen gas are there in the sample.

4 0
3 years ago
The tabulated data were collected for this reaction:
erastovalidia [21]

Answer:

ai) Rate law,  Rate = k [CH_3 Cl] [Cl_2]^{0.5}

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)

If A + B \rightarrow C + D, the rate of backward reaction is given by:  

Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1

Using the information provided in lines 3 and 4 of the table and insering the value of a:

0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\

0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5

The rate law is: Rate = k [CH_3 Cl] [Cl_2]^{0.5}

The rate constant k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}} then becomes:

k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0
3 years ago
Please help will reward brainliest
dimaraw [331]
The answer is most definitely “A”
7 0
3 years ago
(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati
masya89 [10]

Answer:

Explanation:

The solution has been attached

3 0
3 years ago
Several billion years ago, which gas was the least prevalent in the atmosphere?
Amiraneli [1.4K]
The least prevalent gas in the atmosphere was oxygen several billion years ago. This is because it was only around 2.4-3 billion years ago when the first photosynthetic bacteria evolved, meaning they produce food from inorganic compounds such as carbon dioxide and water to produce glucose and oxygen. Oxygen is then released in the atmosphere and this is called <em>The Great Oxygenation Event</em>. 
7 0
3 years ago
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