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olga_2 [115]
3 years ago
6

The decay of which radioisotope can be used to estimate the age of the fossilized remains of an insect?

Chemistry
2 answers:
exis [7]3 years ago
4 0

Answer:

C-14

Explanation:

Carbon-14 is a naturally occurring isotope of carbon.

It is present in all living organism.

When a living organism dies or fossilized, the radioactive carbon-14 undergoes disintegration.

If we can measure the amount of Carbon-14 left in a fossil we can determine the age of the dead organism.

This is known as carbon dating.

Dmitriy789 [7]3 years ago
3 0
(4) C-14 Carbon is found in all living organisms.
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Burning one gallon of gasoline releases 20 pounds of CO2 into the air. Harry’s compact car can travel 25 miles on one gallon of
son4ous [18]
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
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So 
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
                                                      = 12000 pounds
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5 0
3 years ago
What is the precipitation reaction of sodium nitrate and lead (II)?
Vadim26 [7]
The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.
8 0
3 years ago
How might cell division help multicellular organisms maintain and repair cells
Komok [63]
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4 0
3 years ago
Opal is a hydrated form of silica. If a laboratory analysis of a sample of opal reveals it to contain29.2% Si, 33.3%O, and 37.5%
ZanzabumX [31]
Let us assume that there is a 100g sample of Opal. The masses of each element will be:
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33.3g O
37.5g H2O
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Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
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O: 2
H2O: 2
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8 0
3 years ago
How many molecules (not moles of nh3 are produced from 5.01×10?4 g of h2?
Irina-Kira [14]
2,02g     -------    6,02×10²³
5,01×10⁴g  ---    x

x=\frac{5,01*10^{4}g*6,02*10^{23}}{2,02g}=14,93*10^{27}

N₂         +        3H₂       ⇒           2NH₃
1mol      :        3mol        :           2mol
                       18,06×10²³  :       12,04×10²³
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y=\frac{14,93*10^{27}*12,04*10^{23}}{18,06*10^{23}}\approx9,95*10^{27}
5 0
3 years ago
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