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11111nata11111 [884]
2 years ago
8

A coin collector has a bag of 50 pennies. What is the percent abundance of the post-1982 pennies? ASAP ANSWER PLS.

Chemistry
1 answer:
Novosadov [1.4K]2 years ago
5 0

The percent abundance of the post-1982 pennies is 56%.

<h3>What is percent abundance?</h3>

Percent abundance is the percentage amount of all naturally occurring isotopes of an element.

<h3> Percent abundance of the post-1982 pennies</h3>

The percent abundance of the post-1982 pennies is calculated as follows;

percent abundance = number of  post-1982 pennies / total number of pennies x 100%

percent abundance =  (28) / (50) x 100%

percent abundance =  56%

Thus, the percent abundance of the post-1982 pennies is 56%.

Learn more about percent abundance here: brainly.com/question/6844925

#SPJ1

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1. In general, the nucleus of a small atom is stable. Therefore, over very short distances, such as those in a small nucleus,
postnew [5]
<h3>1. <u>Answer;</u></h3>

a. the strong nuclear force is much greater than the electric force.

<h3><u>Explanation</u>;</h3>
  • <em><u>For an atom to be stable it means it has enough amount of binding energy to hold its nucleus together permanently. </u></em>
  • Therefore, <em><u>an unstable atom lacks enough amount of binding energy to hold its nucleus permanently and thus undergoes decay to achieve stability. Unstable atoms are therefore referred to being radioactive.</u></em>
  • Small atoms are stable; <u>this is because they have equal number of protons and neutrons and thus the protons and neutrons fill up energy levels while maximizing the strong force binding the nucleus together. </u>

<h3>9.<u> Answer;</u></h3>

b. change into a different element altogether.

Uranium-238 undergoes alpha decay. Therefore, uranium-238 will  <em><u>change into a different element altogether</u></em>.

<h3><u> Explanation;</u></h3>
  • Unstable atoms undergo radioactive decay in order to achieve stability of their nucleus.
  • <em><u>Uranium-238 is an example of such atom, which may undergo decay to achieve stability.</u></em>
  • <em><u>Alpha decay is one of the types of decays,</u></em> others being beta decay and gamma decay. <em><u>In alpha decay the radioactive isotope undergoes decay such that its mass number is decreased by four and its atomic number is decreased by two.</u></em>
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3 years ago
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Which of the following is not a phase change?
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C. Diffusion is not a phase change.
6 0
3 years ago
How many grams are in 4.07x10^15 molecules of calcium hydroxide
maksim [4K]

There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

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2 years ago
For alkyl halides used in SN1 and SN2 mechanisms, rank the leaving groups in order of reaction rate. You are currently in a rank
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Explanation:

During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.

This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).

In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.

Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.

Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride

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