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nalin [4]
3 years ago
6

Which of the following elements could form a diatomic molecule?

Chemistry
1 answer:
olga2289 [7]3 years ago
3 0
C) Fluorine
For a diatomic molecule remember BrINCl HOF
Br- Bromine
I- Iodine
N-Nitrogen
Cl- Chlorine
H- Hydrogen
O-Oxygen 
F- Fluorine 
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The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver
Evgen [1.6K]

Answer: Partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   P_{N_{2}} = x_{N_{2}}P

where,    P_{N_{2}} = partial pressure of N_{2}

                 P = atmospheric pressure

            x_{N_{2}} = mole fraction of N_{2}

Putting the given values into the above formula as follows.

      P_{N_{2}} = x_{N_{2}}P

    593 mm Hg = x_{N_{2}} \times 760 mm Hg

       x_{N_{2}} = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of N_{2} is as follows.

         P_{N_{2}} = x_{N_{2}}P

                  = 0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}

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8 0
3 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
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Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

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

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

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