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LUCKY_DIMON [66]
3 years ago
15

What term defines the specific amount of time required for half of a radioactive substance to become stable?

Chemistry
2 answers:
Genrish500 [490]3 years ago
6 0
A. half-life that is the correct answer because i"ve seen this question before. 

goldenfox [79]3 years ago
5 0
Half life is the specific amount of time required for half of a radioactive substance to become stable.

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C2h4 3 o2 2 co2 2 h2o
Marat540 [252]
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole=  32.00 of O2

O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving

0.624 mol O2    /  1mol of CO2
___________  / _____________ = Cancel the queal ( O2)
                       / 32.00c O2
                      /
                     /
Multiply the top and divide by the bottom 
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there) 
Ur answer should have the same # as ur givin so 
= 0.0195 
= .0195 mol of CO2
3 0
3 years ago
Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re
lapo4ka [179]
a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K 

when ΔG° is Gibbs free energy 

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 =  298 K 

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

      = -7739 J  = -7.7 KJ


b) then, to get E
° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

              = - 2.7 x 10^-2 V
  
8 0
3 years ago
HURRY PLEASE
katrin2010 [14]

Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

<em>Melting point</em> is the temperature at which the substace changes its state from<em> solid to liquid</em>. During that change, <em>the temperature</em> of the substance <em>does not change</em>, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means <em>that portion of the diagram is flat</em>.

The same is valid during<em> boiling</em>: the temperature remains constant while the substance is passing<em> from liquid to gas</em> at the boiling point.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Between the <em>melting</em> (-183°C) and<em> boiling</em> (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a <em>diagonal</em> (slant) <em>line</em> between those points. During this interval the heat is used to <em>increase the temperature</em> and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

<u><em></em></u>

The table shows that the<em> boiling point</em> of gold is 2,856°C.

In a <em>temperature-vs.-time diagram</em> the<em> temperature is represented on the vertical axis (y-value)</em> and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during <em>boiling,</em> the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the <em>top horizontal line in the diagram</em>.

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

<u><em></em></u>

The table shows that the <em>melting point </em>of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

<u>In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the </u><em><u>temperature - vs. - time diagrams </u></em><u>show flat lines (constant y-values) during those intervals of time.</u>

4 0
3 years ago
Read 2 more answers
A sample of hydrogen gas will behave most like an ideal gas under conditions of?
fenix001 [56]
Conditions:
Low pressure and low temperature
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High pressure and low temperature
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6 0
3 years ago
Which organism is a tertiary consumer?<br> clownfish<br> brittle star<br> killer whale<br> octopus
vlada-n [284]

Answer:

killer whale

Explanation:

7 0
3 years ago
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