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Margarita [4]
2 years ago
7

!DUE SOON PLEASE HELP!

Chemistry
1 answer:
nalin [4]2 years ago
6 0

Answer:

balanced;

unbalanced;

unbalanced;

balanced;

Explanation:

CaO + 3C → CaC2 + CO  //all elements are balanced

Na + H2O → 2NaOH + H2  //Na is not balanced on the left

4Fe + O2 → 2Fe2O3  //O is not balanced on the left

2Mg + O2 → 2MgO //left is equal to right, balanced

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Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.​
adelina 88 [10]

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is O_1H_1

and written as OH.

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2 years ago
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Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

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3 years ago
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