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ad-work [718]
3 years ago
13

Analysis of Potassium-40 / Argon-40 determines that 37.5% of the Potassium-40 remains and 62.5% has decayed to Argon-40. What is

the age of the rock? A. 1.3 billion years B. 0.65 billion years C. 1.95 billion years D. 0.163 billion years
Chemistry
1 answer:
likoan [24]3 years ago
6 0

Answer:

Option C is correct.

t = 1.95 billion years.

Explanation:

Radioactive decay follows a first order reaction kinetics.

On solving the dynamic equation (the differential equation), this is obtained

C(t) = C₀ e⁻ᵏᵗ

C(t) = amount of radioactive material remaining after time t = 37.5%

C₀ = Initial amount of radioactive material = 100%

t = time that has passed = ?

k = decay constant.

For a first order reaction, the decay constant is related to the half life through the relation

k = (In 2)/T

T = half life = 1.38 billion years

k = (In 2)/1.38

k = 0.5023 per billion years.

C(t) = C₀ e⁻ᵏᵗ

0.375 = e⁻ᵏᵗ

e⁻ᵏᵗ = 0.375

In e⁻ᵏᵗ = In 0.375 = -0.981

-kt = -0.981

t = (0.981/0.5023) = 1.95 billion years.

Hope this Helps!!!

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If the aluminum block is initially at 25 ∘C∘C, what is the final temperature of the block after the evaporation of the alcohol?
Effectus [21]

Answer:

Final temperature of aluminum block = 12.1°C

<em>Note: The question is not complete. The complete question is given below:</em>

<em>If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.</em>

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol =  0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C

5 0
3 years ago
0.0000250 m in scientific notation
malfutka [58]
2.50 × 10 ^-5. Move your decimal until you get a whole number. Then count how many times you moved it and that is your exponent. So your exponent is -5 because you're going to the left, if you're going to the right it would be a positive number. 
3 0
3 years ago
Question 28 (2 points)
Afina-wow [57]

Answer:

Il of the above are part of the Kinetic Molecular Theory

Explanation:

4 0
3 years ago
Calculate the amount of heat needed to boil 41.1 g of water (H2O), beginning from a temperature of 84.7 C . Be sure your answer
Levart [38]

Explanation:

We need to go through to stages to boil 41.1 g of water. We have to heat the sample of water from 84.7 °C to 100 °C (the boiling point) And then we have to provide enough heat to boil all the sample of water.

<em>a) Heating from 84.7 °C to 100 °C:</em>

This is calculated using the formula:

Q₁ = m * C * ΔT

Where Q₁ is the amount of heat, m is the mass of the sample, C is the specific heat of water and ΔT is the temperature change. We already know these values:

m = 41.1 g

C = 4.184 J/(g*°C)

ΔT = Tfinal - Tinitial = 100 °C - 84.7 °C

ΔT = 15.3 °C

Replacing these values we can get the amount of heat necessary for the first step:

Q₁ = m * C * ΔT

Q₁ = 41.1 g * 4.184 J/(g°C) * 15.3 °C

Q₁ = 2631 J

<em>b) Boiling 41.1 g of water:</em>

To find the amount of heat that we need to provide to the sample of water to completely boil it we can use this formula:

Q₂ = m * Cv

Where Cv is the latent heat of vaporization.

Cv = 2256 J/g

Q₂ = m * Cv

Q₂ = 41.1 g * 2256 J/g

Q₂ = 92721 J

<em>c) Total amount of heat:</em>

Qtotal = Q₁ + Q₂

Qtotal = 2631 J + 92721 J

Qtotal = 95352 J = 95400 J

Qtotal = 95.4 kJ

Answer: The amount of heat needed to boil the sample of water is 95.4 kJ or 95400 J.

8 0
2 years ago
Worth 40 Points❤ 10th grade chemistry ‍
brilliants [131]

Answer:

  1. F
  2. E
  3. G
  4. A
  5. C
  6. B
  7. D

Explanation:

Those are the answers in order, BUT the Goldstein and Rutherford ones are confusing me because Rutherford discovered<em> protons</em> and the <em>nuclear atom</em>. Through my research I also noticed that Goldstein contributed to the discovery of the protons made Rutherford, so I listed him as the one who discovered the proton.

I hope this helps! Have a wonderful night! :D

6 0
3 years ago
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