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Oksanka [162]
3 years ago
14

What’s the correct answer

Chemistry
1 answer:
Fed [463]3 years ago
3 0
The correct answer is: 
_________________________________________________________
    " F  and Br , because they are in the same group" .
_________________________________________________________
Note:
_________________________________________________________
Choice [B]:  "F and Br ; because they are in the same period" ; is incorrect; since  "F" and "Br" are not in the same "period" (that is, "row").
______________________________________________________
Choice [C]:  "Na and Mg; because they are in the same group {"column"} ;  is incorrect; since:  "Na" and "Mg are NOT in the same group {"column"].
_______________________________________________________
Choice [D]:  "Na and Mg" ; because they are in the same period {"row"}; is incorrect.  Note:  "Na" and "Mg" are, in fact,  in the same period {"row"}. However, as aforementioned, {Mg and Na} are not in the same group {"column".}.  

Note:  The similiarities in physical and chemistry properties among elements are determined and organized — or tend to be so—by "groups" {"columns"} — NOT by "periods" {rows}.  
______________________________________________________
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Scorpion4ik [409]
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4 years ago
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Min : sec helium and nickel are classified as _____, which are pure substances that make up all kinds of matter
Sidana [21]

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4 0
3 years ago
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)&#10;

Rate Law: k[HI]^2&#10;

Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
4 years ago
CAN SOMEONE PLEASE HELP MEEE :(
o-na [289]

Answer:

B

Explanation:

The water is the solvent and the sugar is the solute

Hope this helps!

5 0
3 years ago
PLEASE HELP ITS DUE TODAY!
Yuri [45]

Answer:

At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)

5 0
2 years ago
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