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lilavasa [31]
3 years ago
12

How many moles of copper are formed when 3.8 moles of zinc are mixed with copper (II) sulfate?

Chemistry
1 answer:
Musya8 [376]3 years ago
6 0
CuSO₄+ Zn = Cu + ZnSO₄

1 mole Zn ---------> 1 mole Cu
3.8 mole Zn ------> ?

moles Cu = 3.8 x 1 / 1

= 3.8 moles of Cu

hope this helps!
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g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i
kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is  \alpha_{Y^{-4}} =0.81

So the Conditional Formation constant= K_f =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL  

V e= 25.00 mL  

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.  

[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M

                                                            {K^'}_f

                       Ca²⁺      +      Y⁴          ⇄     CaY²⁻

Initial                 0                  0                      0.001667

change             +x                  +x                     -x

equilibrium        x                    x                    0.001667 - x

{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\

x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M  

pca = —log(2 15x101= 6.7

3 0
4 years ago
Given the unbalanced equation: al + cuso4 --> al2(so4)3 + cu. when properly balanced, the sum of the balancing coefficients i
elena-14-01-66 [18.8K]
2Al + 3CuSO₄  =  Al₂(SO₄)₃ + 3Cu

∑=2+3+1+3=9

c. 9
3 0
3 years ago
Whats The amount of space an object takes up is known as the objects?
Ray Of Light [21]
The amount of space would be consider matter
7 0
3 years ago
A ball has a 17 j of kinetic energy and its mechanical energy is 25 j. Calculate its potential energy
ycow [4]

The potential energy of the ball is 8 J.

<h3>What is mechanical energy?</h3>

Mechanical energy is given as the energy possessed by an object to do work. It is given as the sum of the kinetic energy and potential energy.

Mechanical energy = Kinetic energy + Potential energy

The given ball has:

  • Mechanical energy = 25 J
  • Kinetic energy = 17 J

The potential energy is given as;

25 J = 17 J + Potential energy

Potential energy = 25 J - 17 J

Potential energy =  8 J

The potential energy of the ball is 8 J.

Learn more about potential energy, here:

brainly.com/question/24284560

#SPJ4

3 0
2 years ago
A sample of a gas (1.50 mol) is contained in a 15.0 l cylinder. the temperature is increased from 100 °c to 150 °c. what is the
pav-90 [236]

<u>Given:</u>

Moles of gas, n = 1.50 moles

Volume of cylinder, V = 15.0 L

Initial temperature, T1 = 100 C = (100 + 273)K = 373 K

Final temperature, T2 = 150 C = (150+273)K = 423 K

<u>To determine:</u>

The pressure ratio

<u>Explanation:</u>

Based on ideal gas law:

PV = nRT

P= pressure; V = volume; n = moles; R = gas constant and T = temperature

under constant n and V we have:

P/T = constant

(or) P1/P2 = T1/T2 ---------------Gay Lussac's law

where P1 and P2 are the initial and final pressures respectively

substituting for T1 and T2 we get:

P1/P2 = 373/423 = 0.882

Thus, the ratio of P2/P1 = 1.13

Ans: The pressure ratio is 1.13


7 0
3 years ago
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