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Phoenix [80]
2 years ago
13

60 POINTS

Chemistry
1 answer:
Juli2301 [7.4K]2 years ago
5 0

Answer:

PBr3 - Molecule , Polar

N2H2 - Molecule , (Polar in E- form and Non- polar in Z form)

C2H2 - Molecule , Non- Polar

N2 - Molecule , Polar

NCl3 - Molecule , Polar

SiF4 - Molecule , Non- Polar

NH3 - Molecule , Polar

F - Not- Molecule (atom)

H2 - Molecule and Non- Polar

Explanation:

Molecule : these are group of two or more atoms joined by strong force of attraction.

H2  is non- polar because it is homoatomic molecule.(made up of same element)

N2 is non- polar because it is homoatomic molecule.

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2 years ago
The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
Natasha_Volkova [10]

Answer:

Ok:

Explanation:

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We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

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2 years ago
Which of these scenarios involve a reaction that is at equilibrium
Tasya [4]
Reaction is producing more reactants than products
3 0
3 years ago
Read 2 more answers
Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the
Alecsey [184]

Explanation:

The chemical equation is as follows.

      \frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)

And, the given enthalpy is as follows.

    \frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g);  \Delta H = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   \Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}

    102.5 = [(\frac{1}{2})x + 498] - [(2)(243)]

    102.5 = (\frac{1}{2})x + 498 - 486

     102.5 - 12 = \frac{x}{2}

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       \Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})

              x = [(\frac{1}{2})181 + (\frac{1}{2})498] - 243

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0
3 years ago
The nucleus’ pull on its electrons, taking into account any attractions weakened by shielding is called the:
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Answer:

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