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gogolik [260]
4 years ago
14

The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia

l volume of 49.16±0.06 mL was delivered, the buret was refilled, and an additional 1.69±0.04 mL was delivered before the endpoint was reached. The titration of a blank solution without the analyte required 0.33±0.04 mL . Calculate the endpoint volume corrected for the blank and its absolute uncertainty. Note: Significant figures are graded for this problem. To avoid rounding errors, do not round your answers until the very end of your calculations.
Chemistry
1 answer:
Margaret [11]4 years ago
7 0

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

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3 years ago
Which of the following best describes a decomposition reaction?
Ilia_Sergeevich [38]
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3 years ago
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Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

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6 0
4 years ago
F(x) = x2. What is g(x)?  A.   B. g(x) = (16x)2  C. g(x) = (4x)2  D. g(x) = 4x2
Alona [7]

D. g ( x ) = 4x^2

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8 0
3 years ago
To a 25.00 mL volumetric flask, a lab technician adds a 0.250 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
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Answer: The number of moles of the weak acid in the solution is 0.00396

Explanation:

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The balanced chemical reaction is:

HA+KOH\rightarrow KA+H_2O  

1 mole of base uses = 1 mole of HA  

0.00396  moles of base use = \frac{1}{1}\times 0.00396=0.00396 moles of HA  

Thus the number of moles of the weak acid in the solution is 0.00396

6 0
3 years ago
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