Calculate the empirical formula for DCBN, given that the percent composition is 48.9%C, 1.76%H, 41.2%Cl, and 8.15%N.

Chemistry

1 answer:

sweet [91]3 years ago

5 0

Answer:

empirical formula is C7H3NCl2

Explanation:

too much work too explain and im lazy

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Viefleur [7K]

The answer is: True.

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likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

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madreJ [45]

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The molar mass of sodium carbonate is 105.9888 g/mol (grams per mole)

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