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Zepler [3.9K]
3 years ago
8

A substance that does not dissolve in a solvent is said to be:

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0
A is the I think 83388
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Red blood cells contain an enzyme called carbonic anhydrase, which catalyzes the reversible reaction whereby carbon dioxide and
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Determine total H for bonds broken and formed, the overall change in H, and the final answer with units. Is it ENDOthermic or EX
Mrac [35]
  • E(Bonds broken) = 1371 kJ/mol reaction
  • E(Bonds formed) = 1852 kJ/mol reaction
  • ΔH = -481 kJ/mol.
  • The reaction is exothermic.
<h3>Explanation</h3>

2 H-H + O=O → 2 H-O-H

There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb

  • E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
  • ΔH(Breaking bonds) = +1371 kJ/mol

Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release

  • E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
  • ΔH(Forming bonds) = - 1852 kJ/mol

Heat of the reaction:

  • \Delta H_{\text{rxn}} = \Delta H(\text{Breaking bonds}) + \Delta H(\text{Forming bonds})\\\phantom{ \Delta H_{\text{rxn}}} = +1371 + (-1852) \\\phantom{ \Delta H_{\text{rxn}}} = -481 \; \text{kJ} / \text{mol}

\Delta H_{\text{rxn}} is negative. As a result, the reaction is exothermic.

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F liquid water runs into an open space within a rock, like a crack, and the temperature drops to the freezing point of water, wh
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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

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3 years ago
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