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noname [10]
2 years ago
15

How to balance this equation: Cu(s) + O2(g) --> CuO(s)

Chemistry
1 answer:
Lelu [443]2 years ago
3 0
2Cu(s) + O2(g) --> 2CuO(s)
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Mechanism with a small activation energy or one with large activation energy​
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Answer:

Rate depends on the rate constant. The rate constant depends on temperature and activation energy. If you have lower activation energy the rate will be higher. This is why catalysts are added since catalysts provide an alternate pathway that requires lower activation energy and catalysts are added to increase the rate of reaction.

Explanation:

This is only the answer if you were asking:

"Which corresponds to the faster rate: a mechanism with a small activation energy or one with a large activation energy?"

Thats what I understood about your question.

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Which sample would serve as a better buffer and therefore a better environment for aquatic life?
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sample A

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the first one because of the ppm value

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What is the element with 2 protons on the
Paul [167]

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Helium

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Is it called when a substance enters a gaseous phase without boiling?
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In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul
BARSIC [14]

<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m = molality of barium iodide solution = 0.175 m

m_{solute} = Given mass of solute (barium iodide) = 13.9 g

M_{solute} = Molar mass of solute (barium iodide) = 391.14 g/mol

W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g

Hence, the mass of water that should be added in 203.07 grams

4 0
3 years ago
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