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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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a cylinder container is filled with air. its radius is 6cm and its height is 3.5cm. what is the volume of the air inside?
hram777 [196]

Answer:

45

Explanation:

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4 0
3 years ago
Fill in the blanka gas-like state of matter consisting of free electrons and atomic nuclei is __________
belka [17]
Hi lovely,

The answer you're looking for would be Plasma.
7 0
3 years ago
Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/
Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of C_2H_4 and H_2 = 25.0 atm

Temperature of C_2H_4 and H_2 = 250^oC=273+250=523K

Volume of C_2H_4 = 1050 L per min

Volume of H_2 = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of C_2H_6 = 30 g/mole

First we have to calculate the moles of C_2H_4 and H_2 by using ideal gas equation.

For C_2H_4 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}

n=611.34moles

For H_2 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}

n=902.46moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_2H_4+H_2\rightarrow C_2H_6

From the balanced reaction we conclude that

As, 1 mole of C_2H_4 react with 1 mole of H_2

So, 611.34 mole of C_2H_4 react with 611.34 mole of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and C_2H_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C_2H_6.

As, 1 mole of C_2H_4 react to give 1 mole of C_2H_6

As, 611.34 mole of C_2H_4 react to give 611.34 mole of C_2H_6

Now we have to calculate the mass of C_2H_6.

\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6

\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g

The theoretical yield of C_2H_6 = 18340.2 g

The actual yield of C_2H_6 = 14.0 kg = 14000 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of C_2H_6

\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%

Therefore, the percent yield of the reaction is, 76.34 %

5 0
3 years ago
50 points - I need this really soon! plz help
kirill115 [55]

Answer:

C.

Explanation:

Butane (C4H10) has 3 C-C bonds in the carbon chain and 10 C-H bonds

7 0
3 years ago
Identify the Lewis acid and Lewis base in each reaction:Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆
klemol [59]

In the reaction, Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆, Na⁺ is a Lewis acid and H₂O is a Lewis base.

<h3>What is a Lewis acid-base reaction?</h3>
  • According to the Lewis theory of acid-base reactions, acids accept pairs of electrons and bases donate pairs of electrons.
  • Any substance like H+ ion, which is capable of accepting a pair of nonbonding electrons or an electron-pair acceptor is known as a Lewis acid.
  • Any substance, like the OH- ion, that is capable of donating a pair of nonbonding electrons or an electron-pair donor is a Lewis base.
  • Here Na⁺ that is electron deficient accepts electrons from the electron donor, H₂O
  • From the Lewis theory, with no change in the oxidation numbers of any atoms, acids react with bases to share a pair of electrons.

To learn more about Lewis acid-base reactions: brainly.com/question/14861040  

#SPJ4                                                                                              

4 0
2 years ago
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