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UkoKoshka [18]
3 years ago
5

CuO+H₂ → Cu +H₂O redox equation

Chemistry
2 answers:
Agata [3.3K]3 years ago
8 0

Answer:

Cu   H₂O

Explanation:

CuO + H₂  -----> Cu + H₂O

Oxidation : Gaining of oxygen OR Losing of hydrogen

Reduction : Gaining of Hydrogen OR Losing of Oxygen

Here ,

CuO gets reduced [ it loses oxygen ] and become Cu

H₂ gets oxidized  [ it gains oxygen ] and becomes H₂O

Substance that gets oxidised = H₂

Substance that gets reduced = CuO

Also ,

Oxidising agent = substance that gets reduced = CuO

Reducing agent = substance that gets oxidized = H₂

Allisa [31]3 years ago
8 0

Answer:

Cu⁺² + 2e⁻ → Cu

Explanation:

A redox reaction is the half-equation where the element is reduced (gains electrons).

Compounds always have to have oxidation numbers that add to 0.

Oxygen's oxidation number is -2.

Stand alone elements always have an oxidation number of 0.

Reactants

Because Copper's oxidation number + Oxygen's oxidation number = 0, and Oxygen's oxidation number = -2, we know Copper's oxidation number is +2.

H₂ has to have a net oxidation of 0, so Hydrogen's oxidation number is 0.

Products

Copper's oxidation number is 0 because it's a standalone element.

2 * Hydrogen's oxidation number + Oxygen's oxidation number = 0, so that means Oxygen's oxidation number is -2, and Hydrogen's is +1.

Because Copper's oxidation number went down from +2 to 0, that means it gained electrons and therefore was reduced. So your answer is:

Cu⁺² + 2e⁻ → Cu

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Answer:

Explanation:

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Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

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=

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Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

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  • (400, 0.2) ⇒ 400 mb, 0.2 ml
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The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

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