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3 years ago

Which radioactive isotope is most useful for determining the age of mastodont bones found in late Pleistocene sediments?

Chemistry

2 answers:

denis-greek [22]3 years ago

8 0

Answer: carbon-14

Explanation: the radioactive isotope suitable for dating Pleistocene sediments is Carbon-14

andrezito [222]3 years ago

8 0

Answer:

Carbon-14

Explanation:

Carbon dating is also known as radiocarbon dating. Carbon dating is a technique used in the determination of of age(s) of an object. It has been proved that carbon dating is efficient for dating archaeological specimens and fossils. Carbon dating method was developed by Willard Libby(A physicist).

Carbon-14 is an isotope of of carbon. Carbon-14 is radioactive. Carbon-14 is very important in many aspect both medically( to detect ulcer causing bacteria) and in the determination of age of an object. It is also used in radio labelling.

All living things maintain certain amount of carbon-14 while still alive, but when living things die, the amount of carbon-14 decreases exponentially. Carbon-14 is produced by cosmic radiation in the atmosphere.

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The reaction between alcohols and carboxylic acids is called

aleksandrvk [35]

Explanation:

When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along with water). This reaction is called the Fischer esterification.

4 0

2 years ago

Calculate the [oh] for 0.46 Hcl

Bad White [126]

Answer:

2.17 e -14

Explanation:

A strong acid like HCl ionize 100 % in water so [H3O+] = 0.46 M

[OH-] = Kw / [H3O+]

= 1.0 e -14 / 0.46

= 2.17 e -14

3 0

3 years ago

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

3 0

3 years ago

What effect does a catalyst have on a reaccion rate an the products formed in a reaccion

elena-s [515]

A catalyst will speed up the activation energy and therefore speed up the reaction. The products will form fast because of this.

8 0

2 years ago

15.0 mL of an unknown clear liquid is added to a 50 mL graduated cylinder. The mass of the liquid is determined to be 12.7 grams

sveticcg [70]

Answer:

$\boxed {\tt A. \ d=0.85 \ g/mL}$

Explanation:

Density is found by dividing the mass by the volume.

$d=\frac{m}{v}$

The mass of the liquid is 12.7 grams.

We know that 15 mL of this liquid was added to a 50 mL graduated cylinder. Therefore, the volume is 15 mL. The 50 mL is not relevant, it only tells us about the graduated cylinder.

$m= 12.7 \ g\\v= 15 \ mL$

Substitute the values into the formula.

$d=\frac{12.7 \ g}{ 15 mL}$

Divide.

$d=0.846666667 \ g/mL$

Round to the nearest hundredth. The 6 in the tenth place tells us to round the 4 to a 5.

$d \approx 0.85 \ g/mL$

The density of the liquid is about 0.85 grams per milliliter and choice A is correct.

4 0

3 years ago