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Zigmanuir [339]
3 years ago
6

What is a property of every mixture

Chemistry
1 answer:
crimeas [40]3 years ago
5 0

Explanation:

can u post a picture of the question ?

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notka56 [123]
The answer’s in the picture

5 0
3 years ago
Increasing the temperature increase the rate of a reaction by doing which of the following
posledela

Answer:

Increase in temperature increases the average kinetic energy of the reactant molecules, hence lowering the activation energy of the reaction.

8 0
3 years ago
Two substances that a green plant must obtain from its environment to carry on photosynthesis
mafiozo [28]

Answer:

Carbon Dioxide and Water

4 0
3 years ago
A chemistry student weighs out of lactic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to
34kurt

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>

<em />

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

<em>Where 1 mole of the acid reacts per mole of the base.</em>

<em />

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

<h3>28.0mL of the 0.0500M NaOH solution</h3>
8 0
3 years ago
Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlori
Helga [31]

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:

  • pH = pKa + log\frac{[NaOCl]}{[HOCl]}

We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:

  • pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
  • [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
  • pH = 7.54 + log \frac{4}{4}
  • pH = 7.54

By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:

  • pH = 7.54 + log\frac{[NaOCl-HCl]}{[HOCl+HCl]}
  • pH = 7.54 + log \frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}
  • pH = 6.82
6 0
3 years ago
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