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WITCHER [35]
3 years ago
12

Charcoal (burned wood) that was used to make prehistoric drawings on cave walls in france was scraped off and analyzed. the resu

lts were 4 mg carbon-14 (parent isotope) and 60 mg nitrogen (daughter isotope). the half-life of carbon-14 is 5,730 years. how old are the cave drawings?
a. 11,460 years
b. 17,190 years
c. 22,920 years
d. the sample is too old to be analyzed by carbon dating.
Chemistry
1 answer:
Finger [1]3 years ago
6 0
TLDR: the answer is C. 22,920 years.

Half-life describes the amount of time for a radioactive substance to decay to one-half of the original substance’s weight. So, if we had 100g of C-14, after 5,730 years, only 50g remain; after another 5,730 years, only 25g would remain, and so on.

In this problem, we are meant to assume that the original amount of C-14 was 64g, and that, through decay, it forms N-14. We can figure out how many half lives have passed by figuring out how much 4 is out of 64 by dividing 64 by two repeatedly. Each time, count a half life.

64 - 32 (1 HL) - 16 (2 HL) - 8 (3 HL) - 4 (4 HL)

In this problem, 4 half lives have passed. We can now multiply this by the time for one half life to find how many years have passed.

4 x 5,730 = 22,920 years

Approximately 22,920 years have passed since the drawing was created.
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The skeletal structure of an organic compound is an abbreviated representation of its molecular structure, they are quick and easy to draw.

For example, the following image shows the skeletal structure of a compound:

The peaks represent the carbons. We must remember that carbon can have a maximum of 4 bonds.

Now, I will show you how is the structure of this specific compound:

This is ternary alcohol, called 2-methyl-2-butanol. If you see carefully, you will notice that each carbon has 4 bonds. The functional groups present will be OH. The skeletal structure will be:

6 0
1 year ago
A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?​
Shalnov [3]

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

5 0
3 years ago
Explain the two parts of a scientific name?
enot [183]

Answer:The binomial nomenclature system combines two names into one to give all species unique scientific names. The first part of a scientific name is called the genus. The second part of a species name is the specific epithet. Species are also organized into higher levels of classification.

6 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
4 years ago
What happens when the kinetic energy of molecules increases so much that electrons are released by the atoms, creating a swirlin
maxonik [38]
When that happens, you get a plasma — the fourth state of matter.
3 0
3 years ago
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