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Assoli18 [71]
2 years ago
11

1. Create a diagram of your electroplating apparatus (an electrolytic cell). Then submit your drawing with the following terms l

abeled correctly.
• anode
• cathode
• copper strip
• battery
• positive terminal
• negative terminal
• place where oxidation occurs
• place where reduction occurs
• electrolyte solution
• coin (or nail)
• direction of electron flow

Chemistry
1 answer:
Anestetic [448]2 years ago
6 0
A picture of the electroplating apparatus can be found attached. The nail (cathode) is completely submerged and that is where reduction happens. In the other side, the copper strip is (anode) where oxidation happens. The electron flow happens from the anode to the cathode. The positive charge of the battery is attached to the anode while the negative side is attached to the cathode.

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Explanation:

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A student weighed out a 2.055 g sample of a cobalt chloride hydrate, ConClmpH2O, where n, m, and p are integer values to be dete
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The given mass of cobalt chloride hydrate = 2.055 g

A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.

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8 0
2 years ago
How many grams of water can be formed from the reaction of 8.76 grams of H2 with 10.5 liters of O2 (at STP) according to the bal
Kisachek [45]

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16.9g of H₂O can be formed

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Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:

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For a complete reaction of 4.345 moles moles of hydrogen are required:

4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant

Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:

0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.

The mass is -Molar mas H₂O = 18.01g/mol-:

0.938 moles * (18.01g/mol) =

<h3>16.9g of H₂O can be formed</h3>
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3 years ago
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