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2 years ago

Carbon-14 has a half-life of 5715 years. It is used to determine the age of

Chemistry

1 answer:

Radda [10]2 years ago

6 0

The sample of Carbon-14 after 28,575 years=0.001875 mg

<h3>Further explanation</h3>

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Carbon-14 has a half-life of 5715 years, so t1/2=5715 years

A sample today contains 0.060 mg of carbon-14, so No=0.06 mg, then :

$\tt Nt=0.06(\dfrac{1}{2})^{28575/5715}\\\\Nt=0.06(\dfrac{1}{2})^5\\\\Nt=0.001875~mg$

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A spectator ion is (Select all that apply.) a. an ionic component of a reactant that is unchanged by the reaction your eye, b. c

DENIUS [597]

Correct Question:

A spectator ion is (Select all that apply.)

- a piece of french fry contaminating the reaction mixture

- an ionic component of a reactant that is unchanged by the reaction

-in this experiment, nitrate ion

- your eye, carefully watching the progress of the reaction

Answer:

- an ionic component of a reactant that is unchanged by the reaction

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.

Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.

The correct option is therefore the option;

- an ionic component of a reactant that is unchanged by the reaction

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3 years ago

Volume of HCl used 25.0mL 4 l

borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB = 25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

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3 years ago

In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio

Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)

Substituting these values we get

The total amount of caffeine obtained during each extraction

= (4.0×C )+ (2.0×4.6 × C)

= 13.2 C

The amount of caffeine remaining in the aqueous solution is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)

Substituting these values we get

The amount of caffeine remaining in the aqueous solution = 4 × C

The fraction of caffeine remaining in aqueous solution is calculated as

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)

=4.0 C/13.2 C

= 1/3.3.

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3 =0.028

Therefore, the total amount of caffeine extracted

=0.070 × (1-(1/3.3)^3)

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3 years ago

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andriy [413]

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The loudness of sound is depend on the amplitude of the sound wave. That means the more the amplitude, the louder will be the sound and the less the amplitude, the low will be the sound.

The loudness of sound does not depend on the tension and wavelength.

Therefore, the correct answer is, increase amplitude.

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3 years ago

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