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3 years ago

What would the percentage of thorium-232 (parent compound) be in a rock that was dated at 2.8 billions years?

Chemistry

1 answer:

just olya [345]3 years ago

5 0

The half-life of Th-232 is 1.405 × 10¹⁰ years

Time elapsed = 2.8 x 10⁹ years

Equation of radioactive decay:

A = A₀ = (1/2)^ t/t₁/₂

where A₀ is the initial amount, A is the amount after time t, t₁/₂ is the half file

The fraction of thorium-232 that remains in the rock after 2.8 billions years is,

A/A₀ = (1/2) ^ (2.8 x 10⁹/ 1.405 × 10¹⁰) = 0.871

Therefore, the percentage of thorium-232 in the rock that was dated at 2.8 billions year = 87.1%

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Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

7 0

2 years ago

Use the law of conservation of mass to answer the questions. Consider a hypothetical reaction in which A and B are reactants and

shutvik [7]

Answer:

26 grams of D will be produced.

Explanation:

The reaction is given by:

A + B -----> C + D

Mass of A reacted = 21 g

Mass of B reacted = 22 g

Mass of C formed = 17 g

Mass of D formed = m =?

According to law of conservation of mass, the total mass of the reactants used is equal to the total mass of the product formed.

Then:

mass of A reacted + mass of B reacted = mass of C formed + mass of D formed

21 + 22 = 17 + m

m = 26 g

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3 years ago

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

Data:

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

$V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL$

4 0

3 years ago

If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp

emmasim [6.3K]

C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.

heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°

final temp = 21 + 14° = 35°C

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3 years ago

Jack lives in a small city in the Pacific Northwest, where the temperatures are cool all year and the sky is frequently cloudy a

kirill [66]

Answer:

Preocouparcor ,imsm oilp

Explanation:

3 0

2 years ago