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navik [9.2K]
3 years ago
7

Consider the reaction, 2 d(g) + 3 e(g) + f(g) => 2 g(g) + h(g) when h is increasing at 0.64 mol/ls, how quickly is e decreasi

ng? give your answer to 3 decimal places.
Chemistry
1 answer:
bekas [8.4K]3 years ago
8 0
Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g). 
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
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What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?
3241004551 [841]

Answer:  " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

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<u>Note</u>: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

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<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

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 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

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<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

    = 13.49 g Al ;

         →  Round to 3 (Three) significant figures;

         →  Since:  "0.500" has 3 (Three) significant figures:

____________________________

   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

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 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

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3 0
3 years ago
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Explanation:

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= 0.9 M of NaOH.

7 0
3 years ago
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