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3 years ago

List the path of the respiratory system

Chemistry

1 answer:

Lena [83]3 years ago

3 0

nasal cavities (or oral cavity) > pharynx > trachea > primary bronchi (right & left) > secondary bronchi > tertiary bronchi > bronchioles > alveoli (site of gas exchange)

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What is the most basic building block of matter?

Viefleur [7K]

Answer would be A)the atom

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3 years ago

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How many atoms, ions, or molecules are in a 1 molar solution?

Nezavi [6.7K]

Molarity refers to moles/ L. This indicates the concentration of a solution containing solute and solvent.

1 molar solution contains 1 mole of solute in 1 L of solvent.

According to Avogadro's number, 1 mole of a substance contains 6.023 x 10^23 molecules, atoms, or ions.

For this reason, 6.023 x 10^23 atoms, ions, or molecules are in a molar solution.

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2 years ago

Use the following equation to answer the questions below.

notsponge [240]

Answer:

202

Explanation:

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2 years ago

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

2 years ago

Read 2 more answers

15 points + brainliest pls help :(

Komok [63]

Answer:

27 g

Explanation:

M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol

100 mL = 0.1 L solution

1.5 M = 1.5 mol/L

1.5 mol/L * 0.1 L = 0.15 mol C6H12O6

0.15 mol * 180 g/1 mol = 27 g C6H12O6

6 0

2 years ago