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Pavlova-9 [17]
2 years ago
14

Help! Im gonna fail if i dont get the answer.

Chemistry
2 answers:
PolarNik [594]2 years ago
6 0

Answer:

haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Explanation:

yea

jolli1 [7]2 years ago
4 0
Ok I will give brainliest the call back to you get in a few hours and mama mommy mama mama bye mama baby mommy mommy
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A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen
Nezavi [6.7K]
Percent error can be calculated by the difference of the theoretical value and the measured value divided by the theoretical value multiplied by 100 percent. 

% error = 27.26 - 27.2 / 27.26 x100
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n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph
olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

8 0
2 years ago
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