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2 years ago

7

Your body contains vast numbers of carbon atoms. How is it possible that some of these carbons may have been part of the body of

a prehistoric creature?

1 answer:

charle [14.2K]2 years ago

6 0

Carbon is reused through the carbon cycle.

There is carbon in oil, which is made of ancient animals, and this carbon is discharged into the air during ignition.

Carbon is then taken in by plants as CO2, which we then eat.

<u>Explanation</u>:

Carbon is reused through the carbon cycle. There is carbon in oil, which is made of ancient animals, and this carbon is discharged into the air during ignition. Carbon is then taken in by plants as CO2, which we then eat.

Matter carries on as indicated by the guideline of conservation of matter: Under conventional conditions, any matter is neither made nor destroyed but instead is reused again and again. It tends to be changed or recombined, however, it doesn't vanish; everything heads off to someplace.

Carbon particles of my body presumably used to be a piece of the body of a prehistoric creature, as chemical components have been utilized and reused by living animals.

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OlgaM077 [116]

Answer:

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Explanation:

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It demonstrates how optical fibers transmit light signals in high-speed communications.

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7 0

2 years ago

At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same condition

yan [13]

Answer:

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H

-241.8kj -------eqn(1)

H2(g)+ 1/2 O2(g) ------>H2O(l).

Δ H =285.8kj ---------eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) ------>>H2O(g)---------------eqn(3)

But the equation from eqn(2) the eqn does go with vaporization so we can re- write as

H2O ------> H2(g)+ 1/2 O2(g)

Δ H= 285.8kj ---------------eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

5 0

2 years ago

Please help like now please

DENIUS [597]

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7 0

2 years ago

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

LiRa [457]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14

3 0

2 years ago

Identify the precipitate (if any) that forms when nach3coo (also written as nac2h3o2) and hcl are mixed. express your answer as

Karolina [17]

When NaCH3Coo mixed with HCl we will get NaCl and CH3CooH as shown in the following balanced equation:

NaCH3Coo + HCl → NaCl + CH3CooH
so from this equation, we can conclude that is no precipitate because all we get is the acetic acid which found in vinegar and the NaCl which is very soluble so we don't have any precipitate.
so, your answer is no precipitate, no reaction

3 0

2 years ago

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