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Alex
2 years ago
10

Question Details

Mathematics
1 answer:
Sliva [168]2 years ago
7 0

Answer:

Atomic mass unit-

It is represented by ‘amu’. It is equal to the quantity 1/12th mass of an atom of C-12.

Actual mass of one atom of C-12 = 1.9924 x 10-23 gm. = 1.9924 x 10-26 Kg.

1 amu = 1.9924 x 10 -23/12 = 1.66 x 10 -24 gm. = 1.66 x 10 -27 Kg.

Atomic mass of an element = mass of one atom of the element / 1 amu

Actual mass of an element = atomic mass of an element in amu x 1.66 x 10 -24 gm.

For ex-

Actual mass of hydrogen atom = 1.008 x 1.66 x 10 -24 gm. = 1.6736 x 10-24 gm.

Actual mass of oxygen atom = 16.00 x 1.66 x 10 -24 gm. = 2.656 x 10-23 gm.

Average Atomic weight (Mass)-

“The Average Atomic Masses of many elements are determined by multiplying the atomic mass of each isotope by its fractional abundance and adding these values and then dividing it by 100.”

Ex.- Naturally occurring carbon contains three isotopes – C12 (98.892 % abundance), C13 (1.108 % abundance), C14 (2 x 10-10 % abundance).The relative atomic masses of these isotopes are 12.000, 13.00335 and 14.00317 amu respectively.

Average atomic mass = % of I isotope x its atomic mass + % of II isotope x its atomic mass +% of III isotope x its atomic mass/100

= 12 x 98.892 + 13.00335 x 1.108 + 14.00317 x 2 x10-10 /100

Average atomic mass= 12.011 amu

Molar volume –

Molar volume of a substance is volume occupied by 1 mole of that substance .

Molar volume of solid or liquid = molar mass / density

Molar volume of ideal gas at 00C or 273 K and 1 atmosphere pressure is 22.4 litre.

Gram atomic mass(weight) or gram atom –

“Gram atomic mass(weight) of an element is the mass of Avogadro number ( 6.023 x 1023 ) of atoms of that elements in grams.”

Ex –

mass of 1 atom of oxygen = 16 amu = 16 x 1.66 x 10-24 gm

Mass of 6.023 x 1023 atoms of oxygen= 16 x 6.023 x 1023 x 1.66 x 10-24 = 16 gm.

Hence, Gram atomic weight of oxygen = 16 gm.

No. of gram atoms = mass of element in gm / atomic mass of element in gm.

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Step-by-step explanation:

The given system of equation is

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On comparing Matrix  A_1 with Matrix A

\begin{bmatrix}\\a &c \\ \\b &d \\\end{bmatrix}=\begin{bmatrix}\\2&8 \\ \\4 &-2 \\\end{bmatrix}

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And if we solve for a we got

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So the value of height that separates the bottom 20% of data from the top 80% is 23.432.  

Step-by-step explanation:

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

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If we use condition (b) from previous we have this:

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But we know which value of z satisfy the previous equation so then we can do this:

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