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agasfer [191]
2 years ago
14

If the pOH of a nitric acid (HNO3) solution is measured as 11, what is the concentration of HNO3 solution?

Chemistry
1 answer:
Illusion [34]2 years ago
4 0

1 gram of HNO3 will be equal to 1/63.01 moles. Therefore, we can say that 1 liter of Nitric acid contains 15.6976 moles or in other words molarity of 70% (w/w) Nitric acid is equal to 15.6976 M.

Known values  

Molecular weight of HNO3 63.01 g/mole

Concentration of Nitric acid 70% (% by mass, wt/wt)

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

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the answer you are looking for is 4.22 kj.

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In organic chemistry, a reaction in which the Grignard reagents or organometallic substances are added to organic compounds such as aldehydes and ketones to form alcohol is known as Grignard reaction.

These Grignard reagents are magnesium halides of alkyl, vinyl or allyl, which react with a carbonyl group to form alcohols.

During this reaction, primary, secondary and tertiary alcohols are formed.

While Biphenyl and benzene are common side products.

These are removed during trituration process in which cold petroleum ether is added to dissolve the biphenyl and benzene side products

If you need to learn more about Biphenyl and benzene click here:

brainly.com/question/4336669

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