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mina [271]
3 years ago
15

What mass of aluminum will be deposited on the cathode if an electric current of 0.15 a is run for 4 5s when a solution of al(no

3 )3 is subjected to electrolysis?
Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Mass of Al, m= ?

Molar mass of copper Al, M = 27.0 g/mol

time, t = 45 s

current , c = 0.15 A

valency of Al in Al(NO₃)₃ , z = +3

From Faraday's 1st law,

m = Mct/zF

where F = Faraday constant = 96500 C/mol

Substitute all the values,

m = [(27.0 g/mol) (0.15 A) (45 s)]/ [3] [96500 C/mol]

= 6.3 x 10⁻⁴ g

Mass of aluminium = 6.3 x 10⁻⁴ g

Therefore, mass of aluminium plated on electrode =  6.3 x 10⁻⁴ g

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The radius of the smaller circle is 8 feet. The distance from the rim of the inner circle to the rim of the outer circle is 3 fe
postnew [5]

Answer:

178.98 sq. feet

Explanation:

The path and the garden has been shown in the figure below. The green area is the garden and the area in brown is the path.

It has been given that,

Radius of garden = 8 feet

So, the area of garden = 3.14 × 8 × 8 = 200.96 sq. feet

The total radius of the land including garden and path = 8 + 3 = 11 feet

So, the total are of land including garden and path = 3.14 × 11 × 11 = 379.94 sq. feet

So, the area of path = Total area of the land - area of garden

Area of path = 379.94 - 200.96 = 178.98 sq. feet

5 0
3 years ago
Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the a
inessss [21]

Write procedural steps that allow you to demonstrate the sun's role in the water cycle using common material - for each explain what you are modeling and how the materials you have chosen represent nature.

Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the accepted value for the density of water

3 0
3 years ago
The Ksp of yttrium iodate, Y(IO3)3 , is 1.12×10−10 . Calculate the molar solubility, s , of this compound.
torisob [31]

Answer:

1.427x10^-3mol per L

Explanation:

Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}

I could use ⇌ in the math editor so I used ----

from the question each mole of Y(IO3)3 is dissolved  and this is giving us a mole of Y3+ and a mole of IO3^3-

Ksp = [Y^3+][IO3-]^3

So that,

1.12x10^-10 = [S][3S]^3

such that

1.12x10^-10 = 27S^4

the value of s is 0.001427mol per L

= 1.427x10^-3mol per L

so in conclusion

the molar solubility is therefore 1.427x10^-3mol per L

3 0
3 years ago
Is a stick of dynamite potential energetic or kinetic
qwelly [4]

Answer:Potential Energy

Explanation:

It is because a stick dynamite is a stored energy which have to be acted upon(lighting it) for it to explode

3 0
2 years ago
Read 2 more answers
What is the volume of 0.80 grams of O2 gas at STP? (5 points) Group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Anarel [89]

Answer:

0.56 liters

Explanation:

First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:

  • 0.80 g ÷ 32 g/mol = 0.025 mol

At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:

  • 0.025 mol * 22.4 L/mol = 0.56 L

Thus the answer is 0.56 liters.

3 0
2 years ago
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