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krek1111 [17]
2 years ago
7

Carbon 14 has a half-life of 5730 years. A geologist has dated a fossil sample, at roughly 28650 years. How much carbon 14 remai

ns in the fossil sample?
the carbon 14 has undergone 5 half-lives.
Chemistry
1 answer:
IgorC [24]2 years ago
7 0
<h3>Answer </h3>

After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be

Explanation:

hope this help

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Explain why energy sources do not have 100% efficiency. Why do you think some have lower efficiencies?
bearhunter [10]

Answer:

Energy sources do not have 100% efficiency because <em>the processes of energy conversion to usable forms involves energy losses. </em>

Some have lower efficiencies due to; <u>energy losses in form of heat</u> during conversion, <u>poor technology applied during conversion</u> of energy and<u> lack of desire equipment</u> to use in the energy conversion system.

Explanation:

The desired form of energy for use is derived from conversion of energy from the source using an energy converter into another form which is usable. The efficiency of the energy converter is calculated as;

л = output energy/input energy

The efficiency of energy is limited to the cost of equipment required for conversion from energy source by the energy converter to a form which is usable. Additionally, because energy sources are scarce, the technology to use in energy  conversion is a factor affecting energy efficiency in that high efficiency will require advanced technology with better equipment leading higher costs of that energy form. when heat losses are involved during energy conversion, efficiency lowers, thus its better if such losses are used as energy input in another system.

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What is the atomic mass number of an isotope that has an atomic number of 5 and includes 6 neutrons?
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The answer should be; 11

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Which is greater 50 degree celcius or 50 degree fahrenheit?with solution plz.....​
zheka24 [161]

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a solution is made by completely dissolving 90. grams of kno3 in 100 grams of water in a beaker. the temperature of this solutio
jolli1 [7]
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On other hand, pressure has a pronounced effect on the solubility of gas in liquid type solutions. In such system, gas is solute (minor component) and liquid is solvent (major component). Example of such solution is aerated water. Herein, CO2 is dissolved in water. In such gas in liquid type of solutions, solubility increases with increasing pressure. 
3 0
3 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
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