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KengaRu [80]
3 years ago
10

The ka of phosphoric acid, h3po4, is 7.6  10–3 at 25 °c. for the reaction h3po4(aq) h2po4 – (aq) + h+ (aq) ∆h° = –14.2 kj/mol.

what is the ka of h3po4 at 60 °c?
Chemistry
1 answer:
blsea [12.9K]3 years ago
7 0
<span>Use the van't Hoff equation: ln ( K2 K1 ) = Δ Hº R ( 1 T1 ⒠1 T2 ) ln ( K2 7.6*10^-3 ) = -14,200 J 8.314 ( 1 298 ⒠1 333 ) ln ( K2 7.6*10^-3 ) = ⒠1708 ( 0.00035 ) ln ( K2 0.0076 ) = ⒠0.598 Apply log rule a = log b b a -0.598 = ln ( e ⒠0.598 ) = ln ( 1 e 0.598 ) Multiply both sides with e^0.598 K 2 e 0.598 = 0.0076 K e 0.598 e 0.598 = 0.0076 e 0.598 K 2 = 0.0076 e 0.598 = 4.2 ⋅ 10 ⒠3 K2 = 4.2 ⋅ 10 ⒠3</span>
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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:  

H₂ (g)  ;   that is:  a "diatomic element" (diatomic gas) ;

_________________________________________

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

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

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Note the conversion for a gas at STP:

______

  1 mol of a gas = 22.4 L gas;

___

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

____

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 ;

____

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 " .

  → as such: We round to "2 (two) significant figures."

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