Please help meee??????

Suppose you are performing a titration. at the beginning of the titration, you read the titrant volume as 2.42 ml. after running

Luda [366]

The volume of titrant required for the titration would be 19.81 mL.

Since the burette was not filled to the zero mark during the titration and the level of base titrant was not filled to the 2.42 mL mark. As a result, the difference between the two values represents the total amount of titrant used in the titration.

therefore,

Volume of titrant after running titration - Volume of titrant before running titration = total titrant required for the titration

22.23 - 2.42 = 19.81 mL

What Exactly Is Titration?

Titration is a laboratory technique that uses a solution with known volume and concentration to determine the concentration of an unknown solution. Between the two solutions, an oxidation-reduction reaction or acid-base neutralization occurs, and the known quantities are used to calculate the unknown. The known concentration standard solution is referred to as the titrant or titrator, while the unknown concentration solution is referred to as the titrand or analyte.

Find more on titration at : brainly.com/question/21881827

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5 0

1 year ago

During a lab investigation Robert was given 5 samples of different metals. Each sample was the same color and had the same volum

Kobotan [32]

He can do this by finding the all the characteristics of copper,
this includes size, density, color,malleable, and see if its a conductor of heat

6 0

3 years ago

No links right answers only!! 9recipitation like in the Arctic tundra.

erik [133]

The answer is. B hope it helps plz mark me a brainless plz

4 0

3 years ago

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

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The diagram below shows a gas sample being measured with an open-end mercury (Hg) manometer. If a barometer reads 730. 1 torr, w

IRISSAK [1]

The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
Change in height (Δh) = 5.89 cm
Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
Pressure of gas (P) =?

<h3>How to determine the pressure of the gas</h3>

The pressure of the gas can be obtained as illustrated below:

P = Pa + PΔh

P = 730.1 + 58.9

P = 789 mmHg

Divide by 760 to express in atm

P = 789 / 760

P = 1.04 atm

Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm

Learn more about pressure:

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Missing part of question:

See attached photo

5 0

1 year ago