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3 years ago

Match the labels to the

Chemistry

2 answers:

Ipatiy [6.2K]3 years ago

7 0

For the first blank, that is the endoplasmic reticulum
For the second, it is lysosome
For the third blank, it is the cell membrane
For the fourth, sorry I don’t know this one
For the fifth, that is the vacuole
For the sixth, that is mitochondrion
For the seventh, that is Golgi body
And lastly the eighth, it is the nucleus
Sorry I did not know what the fourth was but everything else is good.

igomit [66]3 years ago

5 0

Answer: For the first blank, that is the endoplasmic reticulum

For the second, it is lysosome

For the third blank, it is the cell membrane

For the fourth, it is the ribosomes

For the fifth, that is the vacuole

For the sixth, that is mitochondrion

For the seventh, that is Golgi body

And lastly the eighth, it is the nucleus

It goes in a circle from top left to top right

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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

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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 %

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