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Novay_Z [31]
2 years ago
11

Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlig

ht. After one hour, he opens both bottles. Which bottle will likely have more fizzing and bubbles? Why?​
Chemistry
2 answers:
EastWind [94]2 years ago
3 0

Answer:

the soda bottle in the fridge will have more bubbles

Explanation:

MrMuchimi2 years ago
3 0
The soda in the fridge will have more bubbles, because when u leave a carbonated drink in the heat it becomes flat. And the carbonation goes away.
You might be interested in
Please help with 3! Please give only the correct answer...
cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
__________________________________________________
   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
______________________________________________________
4 0
3 years ago
Two of the substances in part 1 are ionic. which factors will result in a stronger ionic bond overall?
olya-2409 [2.1K]
Greater absolute charge 
- This is because ionic bond results from stronger electrostatic forces of attraction.
- The higher the value of charges q₁ and q₂ the stronger will be the ionic bond. 
4 0
3 years ago
Read 2 more answers
Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2
love history [14]

Answer:

<u>Option 1</u>:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

Explanation:

<u>Ionic radius is the radius of an atom´s ion in ionic crystal structure</u>.<u> </u><u><em>In an ion that lose an electron, to form a cation, the radius of the ion gets smaller</em></u><em>, </em>because the repulsion between electrons decrease because fewer electrons are present. Conversely, <u><em>adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.</em></u>                  

<u><em>Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers</em></u>.                        

<u><em>In a isolelectronic series (same number of electrons),</em></u> <u><em>the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius </em></u>beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

<u>We will use this principle to solve our problem</u>.  

In our case, the given ions are:  

  • N⁻³ :    Z = 7,  e⁻ = 10
  • O⁻²:     Z= 8,   e⁻ =10
  • F⁻:       Z = 9,  e⁻ = 10
  • Na⁺:    Z= 11,   e⁻ = 10
  • Mg⁺²:  Z=12,   e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

<em><u>Hence the decreasing order of ionic radius is:</u></em>

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

Have a nice day!

4 0
2 years ago
During which stage of life is a human most dependent on others?
AfilCa [17]

Erm, I think when they are little. When they are just born.

7 0
2 years ago
Read 2 more answers
Please Help!
Whitepunk [10]

We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
</span>

3 0
3 years ago
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