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DanielleElmas [232]
3 years ago
13

Please help, please don’t answer links

Chemistry
1 answer:
lakkis [162]3 years ago
3 0

Answer:

0.093 mole of C₆H₁₂.

Explanation:

We'll begin by calculating the molar mass of C₆H₁₂. This can be obtained as follow:

Molar mass of C₆H₁₂ = (12×6) + (12×1)

= 72 + 12

= 84 g/mol

Finally, we shall determine the number of mole in 7.8 g of C₆H₁₂. This can be obtained as follow:

Molar mass of C₆H₁₂ = 84 g/mol

Mass of C₆H₁₂ = 7.8 g

Mole of C₆H₁₂ =?

Mole = mass / molar mass

Mole of C₆H₁₂ = 7.8 / 84

Mole of C₆H₁₂ = 0.093 mole

Thus, 7.8 g contains 0.093 mole of C₆H₁₂.

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A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L
Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l)  ⇄ N₃⁻(aq) + H₃O⁺(aq)          

The pH of this solution is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq)  ⇄  HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles

Now, the number of moles of HN₃ is:

\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles

Then, the pH of the buffer solution after the addition of HCl is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.    

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0
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Explanation:

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Equilibrium constant, K for a given reaction is the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

Thus, the equilibrium constant, K for the above equation can be written as follow:

K = [H2] [CO] / [HCHO]

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