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andrew-mc [135]
3 years ago
11

Helium on the Moon was found to be 0.420% 2He, 2.75% 3He, and 96.83% 4He. What is the average atomic mass of helium on the Moon?

Chemistry
1 answer:
Tasya [4]3 years ago
6 0

Answer:

Average atomic mass  = 3.9 amu

Explanation:

Given data:

Percent abundance of He-2 = 0.420%

Percent abundance of He-3 = 2.75%

Percent abundance of He-4 = 96.83%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100

Average atomic mass =  0.84 + 8.25 +387.32 / 100

Average atomic mass = 396.41 / 100

Average atomic mass  = 3.9 amu.

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Answer:

See explanation

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The complete ionic equation;

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2 years ago
Be sure to answer all parts.
Liula [17]

A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

Therefore,

16.1 cm² = 16.1 × 1×10⁻¹⁰

16.1 cm² = 1.61×10⁻⁹ km²

Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²

<h3>B. How to determine the cost in cent</h3>

We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:

1 cm² = 0.155 in²

Therefore,

16.1 cm² = 16.1 × 0.155

16.1 cm² = 2.4955 in²

Finally, we shall the determine the cost in centas fo r llow:

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1 in² = 294 cent

Therefore,

2.4955 in² = 2.4955 × 294

2.4955 in² = 734 cents

Thus, the cost of the patch is 734 cents

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The atomic mass unit is presently based on assigning an exact integral mass (in amu) to an isotope of carbon.

<h3>What is the atomic mass?</h3>

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