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TEA [102]
3 years ago
13

Find the no. Of electron involved in the electro deposition of 63.5g of cu from a solution of cuso4

Chemistry
1 answer:
sp2606 [1]3 years ago
8 0
This is a redox reaction, meaning reduction-oxidation reaction. This represents the reaction in one side of the electrode in an electrolysis set-up. First, we find the oxidation number of Cu in CuSO4:

(ox. # of Cu)+ ox.# of S + 4(ox.# of oxygen) = 0
(ox. # of Cu) + (6) + 4(-2) = 0
ox. # of Cu = 2+

CuSO4 ---> Cu + SO42-
Cu2+ + SO42-  ---->  Cu + SO42-
Cu2+ -----> Cu + 2e-   (net ionic reaction)

The stoichiometric equation would be 2 electrons per mole Copper. Copper has a molar mass of <span>63.5 g/mol. Then, it would only need 2 electrons.


</span>
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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
Fill in the coefficients to balance the equation for the chemical reaction that occurs:
Dominik [7]

Answer:

The equation is balanced

Explanation:

NaCl (aq) + AgNO3(aq) ––> AgCl (s) + NaNO3 (aq)

NaCl (aq) + AgNO3 (aq)

Na = 1 , Cl=1 , Ag = 1 , No3= 1

AgCl (s) + NaNO3 (aq)

Ag = 1 , Cl=1 , Na = 1 , No3= 1

8 0
2 years ago
Why do covalent compounds tend to have lower melting points than ionic compounds?
vitfil [10]
Because ionic compounds' strong bonds form network structures, which have a stronger attraction than the covalent compounds which are molecules.
6 0
3 years ago
Read 2 more answers
If the mass of a box is 140 g, and the volume is 8 cm3, then the density of the box = ? socratic.org
Ksivusya [100]

Answer:

17.5 g/cm³

Explanation:

We can solve this particular problem by keeping in mind the <em>definition of density</em>:

  • Density = mass / volume

As the problem gives us both <em>the mass and the volume</em> of the box, we can now proceed to <u>calculate the density</u>:

  • Density = 140 g / 8 cm³
  • Density = 17.5 g/cm³

The density of the box is 17.5 g/cm³.

4 0
2 years ago
Of the predators in the image, which one is most likely a keystone species?
Grace [21]

Answer:

mountian lion

Explanation:

7 0
2 years ago
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