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3 years ago

C

Chemistry

1 answer:

Schach [20]3 years ago

8 0

Answer:

2.74 M

Explanation:

Given data:

Mass of sodium chloride = 80.0 g

Volume of water = 500.0 mL

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will convert the mL into L.

500.0 mL× 1 L /1000 mL = 0.5 L

In next step we will calculate the number of moles of sodium chloride.

Number of moles = mass/molar mass

Number of moles = 80.0 g/ 58.4 g/mol

Number of moles = 1.37 mol

Molarity:

M = 1.37 mol/ 0.5 L

M = 2.74 M

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How many moles of water as a gas can be formed 2.45 L

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Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

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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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1 year ago