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Darya [45]
4 years ago
13

3cuso4 + 2al ------ 3cu + al2(so4)3 in the limiting reagent question above if you have 8.3 g of cuso4 and 20.1 g of al how many

grams of copper can you produce?
Chemistry
1 answer:
Alex17521 [72]4 years ago
8 0
Balanced equation for the reaction is;
3CuSO₄ + 2Al ---> Al₂(SO₄)₃ + 3Cu
Stoichiometry of CuSO₄ to Al is 3:2
Number of CuSO₄ moles present - 8.3 g/ 159.6 g/mol = 0.052 mol
Number of Al moles present - 20.1 g/ 27 g/mol = 0.74 mol
We need to first find the limiting reagent. Limiting reagent is the reactant that is fully consumed, amount of product formed depends on the amount of limiting reactant present.
3 mol of CuSO₄ reacts with 2 mol of Al
Therefore 0.052 mol of CuSO₄ reacts with - 2/3 x 0.052 = 0.034 mol of Al required.
But 0.74 mol of Al present , therefore Al is in excess and CuSO₄ is limiting reactant,
stoichiometry of CuSO₄ to Cu is 3:3 = 1:1
Therefore number of Copper moles formed - 0.052 mol
Then mass of Copper formed  - 0.052 mol x 63.5 g/mol = 3.30 g of Copper is produced

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The answer is "Auger".

Explanation:

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3 years ago
Calculate the energy for the transition of an electron from the n = 5 level to the n = 6 level of a hydrogen atom. E = Joules Is
kramer

Answer:

For an electron to move from a lower energy level to a higher energy , that electron needs to absorb energy sufficient enough to excite it to make the transition. Hence it is an absorption process. The required energy of transition  E = 2.665 x 10⁻²⁰J

Explanation:

Using the Rydberg's equation we can calculate the wavelength of the photon of energy transition as follows:

1/λ = R . (1/nf² - 1/ni²)

where

λ is the required wavelength of the photon needed to be absorbed to excite the electron to transit from level 5 to 6.  

(Note that for the electron to transit to from energy level 5 to 6, the photon would have to fall from level 6 to 5 in order to emit the required energy to excite the electron)

R is the Rydberg's constant 1.097 x 10⁷ m⁻¹

nf is the final level of the photon

ni is the initial level of the photon

1/λ = 1.097 x 10⁷ m⁻¹ (1/5² - 1/6²)

1/λ = 1.3407 x 10⁵ m⁻¹

λ = 7.458 x 10⁻⁶ m

This implies that that is the wavelength of the photon required to excite the electron to transit from energy level 5 to 6. Using the equation below, we can calculate the energy of transition as

E = h.c/λ

where

E is the required energy of transition

h is the Planck's constant (6.626 x 10⁻³⁴ Js)

c is the speed of light (3 x 10⁸ms⁻¹)

λ is the wavelength calculated above

E = 6.626 x 10⁻³⁴ Js  x  3 x 10⁸ms⁻¹/ 7.458 x 10⁻⁶ m

E = 2.665 x 10⁻²⁰J

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3 years ago
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Answer:

The air would still be too thin to breathe. The lack of atmosphere would chill the Earth's surface. Organisms that need air to breathe would die.

Explanation:

Hope it helped!

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3 years ago
What is used up in and stops a chemical reaction?​
hodyreva [135]

Answer:

The limiting reagent is the reactant that is used up completely. This stops the reaction and no further products are made. Given the balanced chemical equation that describes the reaction, there are several ways to identify the limiting reagent.

4 0
3 years ago
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ?C. The initial concentrations of Ni2+ and Zn2+ are
inna [77]

Answer:

a) E = 0.477 V

b) E = 0.502 V

c) 0.02 M = [Ni+2]

d)[Zn+2] = 1.81 M

Explanation:

having the following reactions of each cell:

Zn =⇒  Zn+2   +  2e-    +0.76

Ni+2  +  2e- =⇒   Ni       -0.25

Zn  +  Ni+2  ==⇒  Ni     +    Zn+2     Eo = 0.51

a)

The number of electrons being transferred is 2, therefore n = 2 in the Nernst equation

E =  Eo - 0.0592/2*log [Zn+2]/[Ni+2] = 0.51 - 0.0592/2*log[0.13/1.7] = 0.477 V

b)

using the formula above:

E = 0.51  - 0.0296*log [Zn+2]/[Ni+2] = 0.51 - 0.0296*log((0.13+0.5)/(1.7-0.5)) = 0.502 V

c)

using the formula above:

0.45 = 0.51  - 0.0296*log[Zn+2]/[Ni+2]

-0.06/-0.0296 = log[Zn+2]/[Ni+2]

2.02 = log [Zn+] / [Ni+2]

104.71 = [Zn+2] / [Ni+2]

x = change in [Ni+2]

[Ni+2]  = 1.70 -  x

[Zn+2] =  0.13 +  x

0.13 + x/1.70 –x =  104.71

Resolving  x:

x = 1.68 M

[Ni+2]  = 1.70 -  1.68 = 0.02 M

d)

[Zn+2] =  0.13 +  x = 0.13 + 1.68 = 1.81 M

4 0
3 years ago
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