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asambeis [7]
3 years ago
11

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

to the actual value, and calculates a 25% error. Assuming she has performed all of the calculations correctly, name one aspect of her methods that may have led to such a high error.
Chemistry
1 answer:
Viktor [21]3 years ago
4 0

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

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2 reasons for chemical reactivity of nitrogen
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3 years ago
1) An unlabeled laboratory solution is found to contain a H+ ion concentration of 1 x 10-4
marishachu [46]

Answer:

C) 1 x 10-10 M

Explanation:

To solve this question we must use the equation:

Kw = [H+] [OH-]

<em>Where Kw is the equilibrium dissociation of water = 1x10-14</em>

<em>[H+] is the molar concentration of hydronium ion = 1x10-4M</em>

<em>[OH-] is the molar concentration of hydroxyl ion</em>

<em />

Replacing:

1x10-14= 1x10-4 [OH-]

<em>[OH-] = 1x10-14 / 1x10-4M</em>

<em>[OH-] = 1x10-10 M</em>

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7 0
2 years ago
Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.
babunello [35]

Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

⇒ Ksp = (3S)³ x (S)

⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

Concentration of Li₃PO₄ precipitate = 0.05

<u>Solution A </u>

0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

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The answer is a strike-slip. More specifically a right-lateral strike-slip.

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