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Margarita [4]
3 years ago
7

A compound X has the following percentage composition 66.7% carbon, 11.1% hydrogen and 22.2% oxygen .Calculate the empirical for

mula of X.The relative molecular mass of X is 72 calculate the molecular formula​
Chemistry
1 answer:
Digiron [165]3 years ago
4 0

Explanation:

c. h. o

66.7%. 11.1%. 22.2%

____. ____. ____

12. 1. 16

1.558. 11.1. 1.39. (divide by the smallest)

1. 8. 1

empirical formula=ch8o

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Organisms can either be made of one cell or made of more than one cell
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The statement is true.

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A galvanic cell is set up with two cells containing a cadmium electrode in 0. 250 m cd(no3)2 solution and an iron electrode in 0
Lesechka [4]

The cell potential is 0.01934 V at 25°C.

<h3>What is a galvanic cell?</h3>

A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts the chemical energy generated by spontaneous redox reactions into electrical energy.

The main components of a galvanic cell are

Anode – Oxidation occurs at the anode

Cathode – Reduction occurs at the cathode

Salt bridge – consists of electrolytes that are needed to complete the circuit in a galvanic cell.

Half-cells – It is important as reduction and oxidation reactions are separated into compartments.

External circuit – it helps in Conducting the flow of electrons between electrodes

At cathode,

Cd^{2+}(aq) + 2e^-\rightarrow Cd(s)

At anode,

Fe(s)+2e^-\rightarrow Fe^{2+}(aq)

Overall cell reaction

Fe(s)+Cd^{2+}(aq)\rightarrow Fe^{2+}(aq) + Cd(s)

E^o_{cell} = E^o_{cathode} - E^o_{anode}

          = -0.40 -(0.44) = 0.04 V

Nernst equation is given by

E_{cell} = E^o_{cell} - \frac{2.303RT}{nF} logQ

At 25°C

E_{cell} = E^o_{cell} - \frac{0.0592V}{n} logQ

Here, n = 2 and log Q = log \frac{0.250}{0.050} = 0.699

E_{cell}  = 0.01934 V

The cell potential is 0.01934 V at 25°C

Learn more about galvanic cell:

brainly.com/question/13918634

#SPJ4

3 0
1 year ago
Read 2 more answers
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