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MArishka [77]
3 years ago
12

You wish to prepare 100 mL of 0.600 M HCl from concentrated HCl (12.85 M). You are provided with a 100 mL volumetric flask and a

10 mL graduated pipet with 0.00 line at the top and calibration lines each 0.1 mL. When filled to the 0.00 mL line this pipet would dispense 10.00 mL of solution when drained leaving only the final inside drop. At what volume reading on the pipet should you start dispensing the 12.85 M HCl to have the correct number of moles to give the 0.600 M solution when dispensed into the volumetric flask?
Chemistry
1 answer:
Soloha48 [4]3 years ago
8 0

Answer:

Because we need to dispense 4.7 mL, the volume reading in the pipet is the 5.3 mL line.

Explanation:

First we use C₁V₁=C₂V₂ in order to <u>calculate the required volume of concentrated HCl</u> (V₁):

12.85 M * V₁ = 0.600 M * 100 mL

V₁ = 4.7 mL

<u>So we need to dispense 4.7 mL of the concentrated HCl solution</u>. The mark in the pipet that would contain that volume would be 10.0 - 4.7 = 5.3 mL

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what volume of hydrogen gas is evolved from a reaction between 0.52 g of Na and water? This gas is collected at 20 C and 745mmHg
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that  is  PV = nRT
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V= ?
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6 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

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