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Serggg [28]
1 year ago
13

A substance decays so that the amount a of the substance left at time t is given by: a = a0 ∙ (0.8)t where a0 is the original am

ount of the substance. what is the half-life (the amount of time that it takes to decay to half the original amount) of this substance rounded to the nearest tenth of a year?
Chemistry
1 answer:
Zolol [24]1 year ago
4 0

The half-life of the substance is 3.106 years.

<h3>What is the formula for exponential decay?</h3>
  • The exponential decline, which is a rapid reduction over time, can be calculated with the use of the exponential decay formula.
  • The exponential decay formula is used to determine population decay, half-life, radioactivity decay, and other phenomena.
  • The general form is F(x) = a.

Here,

a = the initial amount of substance

1-r is the decay rate

x = time span

The equation is given in its correct form as follows:

a = a_{0}×(0.8)^{t}

As this is an exponential decay of a first order reaction, t is an exponent of 0.8.

Now let's figure out the half life. Since the amount left is half of the initial amount at time t, that is when:

a = 0.5 a0

<h3>Substituting this into the equation:</h3>

0.5a_{0} = a_{0}×(0.8)^{t}

0.5 = (0.8)^{t}

taking log on both sides

t log 0.8 = log 0.5

t = log 0.5/log 0.8

t = 3.106 years

The half-life of the substance is 3.106 years.

To learn more about exponential decay formula visit:

brainly.com/question/28172854

#SPJ4

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3 years ago
Reacting Solutions of Aluminum Chloride with Potassium Hydroxide
gizmo_the_mogwai [7]

1Al2(SO4)3 + 3ZnCl2 → 2AlCl3 + 3ZnSO4

The coefficients represents moles. There is 1 mole of Aluminum Sulfate, 3 moles of Zinc(II) Chloride, 2 moles of Aluminum Chloride, and 3 moles of Zinc(II) Sulfate.

Now add all the coefficients/moles.

9 is the sum of all the coefficients.

7 0
3 years ago
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In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?
klio [65]

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the <u>cyclohexanone.</u>

See figure 1

I hope it helps!

4 0
4 years ago
The equal areas law is Keplers 2nd law of motion of planetary motion. It states that _____
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An imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time. Thus, the speed of the planet increases as it nears the sun and decreases as it recedes from the sun.

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Q Q 3. (08.02 MC)
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Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

Explanation:

Given: V_{1} = ?,         M_{1} = 0.55 M

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Formula used to calculate the volume of KBr is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.55 M \times V_{1} = 2.50 M \times 100.0 mL\\V_{1} = 455 mL

Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

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