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ipn [44]
3 years ago
12

3.

Chemistry
1 answer:
swat323 years ago
3 0

Answer:

14.7838

Explanation:

Given that;

A student uses a solution that contains 16 g of water to conduct an experiment and at the end of an hour ; the amount have decreased by 3.5%

The mathematically illustration for this is given as

= 16 g ×  3.5%

= 16 *\frac{3.5}{100}

= 0.56  

After an hour; the amount have decreased by 16 - 0.56

= 15.44

after 2 hours the water decreased by another 4.25%

i.e  15.44 * \frac{ 4.25}{100}

⇒ 0.6562

after two hours , the amount would have decreased by:

= 15.44 - 0.6562

= 14.7838

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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

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3 years ago
3.95 g of sugar (C6H12O6) is dissolved in water to make 158 mL of solution. Find the molarity.
12345 [234]

Answer:

[C₆H₁₂O₆] = 0.139 M

Explanation:

Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.

We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.

For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)

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Moles = 3.95 g . 1mol / 180g → 0.0219 mol

We convert mL to L → 158 mL . 1L/1000mL = 0.158L

M = 0.0219 mol / 0.158L = 0.139 M

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