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const2013 [10]
2 years ago
9

Why is electroplating an example of chemical change?

Chemistry
1 answer:
wel2 years ago
7 0

Yes it is true elsectroplating is involves in chemical and physical change but all these is happened on surface of metal by action of the oxidation or reduction.

Chemical change gives by following process as :

Lot of energy of requires,

New substance formed,

And process is irriversiable .

e. g. Rusting of iron,

Photosynthesis,

Burring of kerosene,

Raising cake batter,

Frying egg.

All these examples Shows the chemical change.

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2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2
Margaret [11]

Answer: 6.75 moles

Explanation:

This is a simple stoichiometry proboe. that I would set up like this:

(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)

That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.

Hope this helps :)

7 0
3 years ago
A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.
sweet [91]
The awnser to ur question is B
7 0
3 years ago
Read 2 more answers
Consider the following reaction 2 N2O(g) => 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
stich3 [128]

The given equation: Q + X ---> Products


4 0
2 years ago
How many moles of nitrogen gas would be produced if 4.92 moles of copper(II) oxide were reacted with excess ammonia in the follo
WITCHER [35]

Answer:

4.92/3

Explanation:

ok????????????

3 0
1 year ago
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