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Evgen [1.6K]
2 years ago
5

A J-shaped tube, closed at one end, is depicted in the figure at the right. It contains N2 gas that is trapped by an unknown liq

uid whose density is twice as large as that of mercury. The other end of the tube is open to the air. The value of h is measured to be 10 mm at sea level on a clear day at a temperature of 27°C. The atmospheric pressure is 1.00 atm. Assuming the vapor pressure of the liquid is zero, what is the concentration of the N2 gas trapped inside the J-shaped tube?
Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
5 0

Answer:

the amount of concentration is about 23.9 grams of concentration you might want to add some hydrochloric acid to get the acidity to about 7 or seven depending on what you are doing

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Star_____ has the greatest absolute brightness ​
Nitella [24]

Answer:

Star A would have the greater absolute brightness. This is because absolute brightness finds out the actual brightness of a star at a standard distance from Earth. If Star A is twice as far from Earth as Star B but they still both appear to have the same amount of brightness.

6 0
3 years ago
A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

8 0
3 years ago
Calculate the pHpH of a 0.10 MM solution of HClHCl . Express your answer numerically using two decimal places.
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Answer: The pH 0f 0.10 solution of HCl is 1.00

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pH is calculated by taking negative logarithm of hydrogen ion concentration.

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According to stoichiometry,  

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Thus 0.10 moles of HCl gives =\frac{1}{1}\times 0.10=0.10 moles of H^+  

Putting in the values:

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pH=1.00

Thus pH 0f 0.10 solution of HCl is 1.00

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