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3 years ago

Can someone help me here please i need this :(

Chemistry

1 answer:

podryga [215]3 years ago

7 0

Start by adding the numbers then divide

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This earth system contains water and warms the atmosphere.

sleet_krkn [62]

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3 years ago

Which of the following best describes careers that use chemistry?

telo118 [61]

Answer:

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3 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

3 years ago

Read 2 more answers

A gas sample has volume 3.00 dm3 at 101

gavmur [86]

Answer:

V₂ = 21.3 dm³

Explanation:

Given data:

Initial volume of gas = 3.00 dm³

Initial pressure = 101 Kpa

Final pressure = 14.2 Kpa

Final volume = ?

Solution;

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 Kpa × 3.00 dm³ = 14.2 Kpa × V₂

V₂ = 303 Kpa. dm³/ 14.2 Kpa

V₂ = 21.3 dm³

Explanation:

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3 years ago

39 which statement about the process of dissolving is TRUE

xz_007 [3.2K]

C !!!!!!!!!!!!!!!!!!!!!!

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3 years ago

Can someone help me here please i need this :(​

Can someone help me here please i need this :(