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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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Mamont248 [21]

Answer:16

Explanation:

5 0
3 years ago
Please begging you guys someone help me
BaLLatris [955]

Answer:

See Explanation

Explanation:

10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.

11) The bombardment of U-235 with neutrons leads to the reaction;

U\frac{235}{92}  + n\frac{1}{0}---> I\frac{138}{53}  + Y\frac{95}{39} +3n \frac{1}{0}

Hence

a = 92, b= 95, c= 53

12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.

Hence;

Th\frac{231}{90} -----> e\frac{1}{0} +Ac \frac{231}{89}

4 0
3 years ago
What criteria are used to classify matter?
Naily [24]
Anything that occupies space and has mass.
6 0
3 years ago
Calculate the freezing point and boiling point of a solution containing 14.2 gg of naphthalene (C10H8)(C10H8) in 111.0 mLmL of b
krek1111 [17]

Answer:

The freezing point of solution = -0.34 °C

The boiling point of solution = 82.98 °C

Explanation:

Step 1: Data given

Mass of naphthalene = 14.2 grams

Molar mass naphthalne = 128.17 g/mol

Volume of benzene = 111.0 mL

Density of benzene = 0.877 g/mL

Kf(benzene)=5.12°C/m

Freezing point benzene = 5.5 °C

Kb(benzene)=2.53°C/m

Boiling point benzene = 80.1 °C

Step 2: Calculate mass of benzene

Mass benzene = density * volume

Mass benzene = 0.877 g/mL * 111.0 mL

Mass benzene = 97.3 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = mass naphthalene / molar mass napthalene

Moles napthalene = 14.2 grams / 128.17 g/mol

Moles naphthalene = 0.111 moles

Step 4: Calculate molality

Molality = moles naphthalene / mass benzene

Molality = 0.111 moles / 0.0973 kg

Molality = 1.14 molal

Step 5: Calculate freezing point  of a solution

ΔT = i*kf*m

ΔT = 1 * 5.12 °C/m * 1.14 m

ΔT = 5.84 °C

ΔT = T(pure solvent) − T(solution)

The freezing point of solution = T pure -  ΔT

5.5 - 5.84 = -0.34 °C

The freezing point is -0.34 °C

Step 6: Calculate boiling point  of a solution

ΔT = i*kb*m

ΔT = 1 * 2.53 °C/m * 1.14 m

ΔT = 2.88 °C

ΔT = Tb (solution) - Tb (pure solvent)

The boiling point of solution = T pure +  ΔT

The boiling point of solution = 80.1 °C + 2.88

The boiling point of solution = 82.98 °C

6 0
3 years ago
Calculate the concentration of acetic acid, HAc, and acetate ion, Ac−, in a 0.25M acetate buffer solution with pH = 5.36. "0.25M
Neporo4naja [7]

Answer:

[HAc] = 0.05M

[Ac⁻] = 0.20M

Explanation:

The Henderson-Hasselbalch formula for the acetic acid buffer is:

pH = pka + log₁₀ [Ac⁻] / [HAc]

Replacing:

5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]

3.981 = [Ac⁻] / [HAc] <em>(1)</em>

Also, as total concentration of buffer is 0.25M it is possible to write:

0.25M =  [Ac⁻] + [HAc] <em>(2)</em>

Replacing (2) in (1)

3.981 = 0.25M - [HAc] / [HAc]

3.981 [HAc] = 0.25M - [HAc]

4.981 [HAc] = 0.25M

<em>[HAc] = 0.05M</em>

Replacing this value in (2):

0.25M =  [Ac⁻] + 0.05M

<em>[Ac⁻] = 0.20M</em>

I hope it helps!

7 0
4 years ago
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