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monitta
4 years ago
6

During a lab activity, students prepared two solutions separately at the same temperature. Later they mixed the solutions and th

e temperature dropped. Why?
A:there was no energy input during the chemical reaction
B:energy was absorbed during the chemical reaction
C:energy was released during the chemical reaction
Chemistry
1 answer:
Yuri [45]4 years ago
6 0
The temperature dropped because B. energy was absorbed during the chemical reaction.
If energy was released, the temperature would rise. If there was no energy input, the temperature would stay the same. Since temperature dropped, it means that energy was absorbed.
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victus00 [196]

Answer:

compound is sp3.

Explanation:

4 0
3 years ago
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What mass in grams would 5.7L of hydrogen gas occupy at STP?​
tekilochka [14]

Answer:  The correct answer is:  " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:  

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The molecular weight of "H" is:  1.00794 g ;

   (From the Periodic Table of Elements).

So, the molecular weight of:  H₂ (g)  is:

    " 1.00794 g * 2  = 2.01588 g ; {use calculator) ;

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Note the conversion for a gas at STP:

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  1 mol of a gas = 22.4 L gas;

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i.e. " 1 mol / 22.4 L " ;

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So:     " 5.7 L H₂ (g)  *  \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;

The "L" ("literes" cancel out to "1" ;  since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

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and we are left with:

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 [5.7 * 2.104588 g ] / 22.4  =  ?  g ;

______________________

→ [ 11.9961516  g ] / 22.4 =

          0.53554248214  g ;l

_____________________________

We round this value to:  " 0.54 g " ;

 → since "5.7 L " has 2 (two) significant figures;  

     22.4 is an exact number conversion;

     and "5.7 L" has fewer significant figures than:

    " 2.104588 " ; or:  " 1.00794 " .

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Hope this is helpful.  Wishing you the best in your academic endeavors!

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8 0
3 years ago
Calculate the solubility of zn(oh)2(s) in 2.0 m naoh solution. (hint: you must take into account the formation of zn(oh)2−4, whi
Brilliant_brown [7]
When we have:

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and:

Zn2+ + 4OH- → Zn(OH)4 2-  with Kf = 2 x 10^15
 
by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

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3 years ago
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nabr + cacl2 are the products

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