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nikdorinn [45]
2 years ago
6

The theoretical yield of zinc oxide in a reaction is 486 g. what is the percent

Chemistry
1 answer:
Alex787 [66]2 years ago
7 0

82.09% is the percent yield if 399 g is produced. Hence, option  C is correct.

<h3>What is the percent yield?</h3>

Percent yield is the percent ratio of actual yield to the theoretical yield.

Given data:

Actual yield = 399 g

Theoretical yield = 486 g

%Yield = \frac{Actual \;Yield  \;X  \;100 }{Theoretical  \;Yield}

%Yield = \frac{399 gX  \;100 }{486 g  \;Yield}

= 82.09%

82.09% is the percent yield if 399 g is produced.

Hence, option  C is correct.

learn more about the percent yield here:

brainly.com/question/15464799

#SPJ1

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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
natita [175]

This question provides us –

  • Weight of \bf  KCl is = 47 g
  • Volume, V = 375 mL

__________________________________________

  • Molar Mass of \bf   KCl –

\qquad \twoheadrightarrow\bf  39.0983 \times 35.453

\qquad \twoheadrightarrow\bf 74.5513

<u>Using formula</u> –

\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{47000}{27956.7375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = 1.68117M

\qquad \pink{\twoheadrightarrow\bf Molarity _{(Solution)}  = 1.7M}

  • Henceforth, Molarity of the solution is = 1.7M

___________________________________________

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Scientific explanations are not always based on empirical evidence <br><br> True or false?
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What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29
hram777 [196]

Henderson–Hasselbalch equation is given as,

                                         pH  =  pKa  +  log [A⁻] / [HA]   -------- (1)

Solution:

Convert Ka into pKa,

                                         pKa  =  -log Ka

                                         pKa  =  -log 1.37 × 10⁻⁴

                                         pKa  =  3.863

Putting value of pKa and pH in eq.1,

                                         4.29  =  3.863 + log [lactate] / [lactic acid]

Or,

                   log [lactate] / [lactic acid]  =  4.29 - 3.863

                   log [lactate] / [lactic acid]  =  0.427

Taking Anti log,

                             [lactate] / [lactic acid]  =  2.673

Result:

           2.673 M  lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

6 0
3 years ago
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