How many milliliters of 0.100 M NaOH is needed to titrate 20.0 mL of 0.100 M HNO3 ?

Answers

Answer 1

Therefore, 50.0 m L of 0.200 M NaOH is required to completely neutralize 100.0 mL of 0.100 M HCl

Related Questions

Element A has two isotopes. The first isotope is present 19.52% of the time and has a mass of 265.1.
The second isotope has a mass of 182.27. Calculate the atomic mass of element A. (To two decimals
places)

Answers

The atomic mass of element A is 198.39 (to two decimal places).

To calculate the atomic mass of element A, we need to consider the relative abundance of each isotope and its corresponding mass. The atomic mass is the weighted average of the masses of all the isotopes, taking into account their relative abundance.

Given:

Isotope 1: Relative abundance = 19.52%, Mass = 265.1

Isotope 2: Relative abundance = 100% - 19.52% = 80.48%, Mass = 182.27

To calculate the atomic mass, we multiply the relative abundance of each isotope by its mass and sum up the results.

Atomic mass = (Relative abundance of Isotope 1 * Mass of Isotope 1) + (Relative abundance of Isotope 2 * Mass of Isotope 2)

Atomic mass = (19.52/100 * 265.1) + (80.48/100 * 182.27)

Calculating the values:

Atomic mass = (0.1952 * 265.1) + (0.8048 * 182.27)

Atomic mass = 51.72752 + 146.661296

Atomic mass = 198.388816

Rounding to two decimal places, the atomic mass of element A is 198.39.

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Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative
abundance. If their masses are 81, 115, and 139 respectively, calculate the atomic mass of element
R. (No decimals).

Answers

The average atomic mass of the element R that has three isotopes is 127.805.

How to calculate average atomic mass?

Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element.

The average atomic mass of an element can be calculated by summing up the product of the percent abundance and masses of each isotope as follows;

Average atomic mass of R = (0.0825 × 81) + (0.2671 × 115) + (0.6504 × 139)

Average atomic mass = 6.6825 + 30.7165 + 90.4056 = 127.805

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Element A has two isotopes. The first isotope is present 18.18% of the time and has a mass of
147.99. The second isotope has a mass of 127.76. Calculate the atomic mass of element A. (To two
decimals places)

Answers

The atomic mass of element A, given that the first isotope has abundance of 18.18% and a mass of 147.99, is 131.43 amu

How do i determine the atomic mass of element A?

From the question given above, the following data were obtained:

Abundance of 1st isotope (1st%) = 18.18%Mass of 1st isotope = 147.99Mass of 2nd isotope = 127.76 Abundance of 2nd isotope (2nd%) = 100 - 18.18 = 81.82%Atomic mass of element A=?

The atomic mass of the element A can be obtain as illustrated below:

Atomic mass = [(Mass of 1st × 1st%) / 100] + [(Mass of 2nd × 2nd%) / 100]

Inputting the given parameters, we have:

Atomic mass = [(147.99 × 18.18) / 100] + [(127.76 × 81.82) / 100]

Atomic mass = 26.90 + 104.53

Atomic mass = 131.43 amu

Thus, the atomic mass of element A obtained from the above calaculation is 131.43 amu

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