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3 years ago

Which term identifies the half-reaction that occurs at the anode of an operating electrochemical cell?

Chemistry

1 answer:

Inga [223]3 years ago

4 0

Oxidation reaction.
Cathode on the other hand is reduction.

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When you step out of the bright sunshine into a dark movie theater, your vision gradually adjusts to the relative gloom. This pr

Setler79 [48]

The answer is A. intensity

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2 years ago

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What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?

Natali5045456 [20]

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2 moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2 moles of P requires 5 moles of chlorine

0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

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5 0

1 year ago

Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

7 0

3 years ago

How many grams of NaCl are needed to make 0.800 liter of a 5.00 M solution?

ziro4ka [17]

Answer:

0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.

Given data:

Molarity = 5.00 M

Formula Mass = 58.5 g/mol

Required volume = 0.800 L

To Find;

Mass in gram = ?

Solution:

Formula for calculating mass in gram is given as,

Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values

Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000

Mass in gram = 0.234 g

6 0

3 years ago

Compound formula: MgCl2

umka2103 [35]

Answer:

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2 years ago

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