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11111nata11111 [884]
2 years ago
14

Calculate the number of sucrose molecules in a 75.0 gram sample.

Chemistry
1 answer:
zzz [600]2 years ago
5 0

Answer:

1.32*10^23 molecules

Explanation:

sucrose formula: C12H22O11

molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol

75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11

0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)

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<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)

KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, A^- ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

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  • If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of H^+ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

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Which of the below options is not a type of chemical bond?
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6 0
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K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)

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