Answer:
THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.
Explanation:
Equation:
Al3+ + 3e- -------> Al
3 F of electricity is required to produce 1 mole of Al
3 F of electricity = 27 g of Al
If 18 g of aluminium was used, the quantity of electricity to be used up will be:
27 g of AL = 3 * 96500 C
18 G of Al = x C
x C = ( 3 * 96500 * 18 / 27)
x C = 193 000 C
For 18 g of Al to be produced, 193000 C of electricity is required.
To calculate the current required to produce 193 000 C quantity of electricity, we use:
Q = I t
Quantity of electricity = Current * time
193 00 = I * 1.50 * 60 * 60 seconds
I = 193 000 / 1.50 * 60 *60
I = 193 000 / 5400
I = 35.74 A
The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A
Answer is: mass of the mixture is 24,47 g.
1) N(Na₂SO₄) = 3,5·10²².
n(Na₂SO₄) = 3,5·10²² ÷ 6·10²³ 1/mol.
n(Na₂SO₄) = 0,058 mol.
m(Na₂SO₄) = 0,058 mol · 142 g/mol.
m(Na₂SO₄) = 8,24 g.
2) n(H₂O) = 0,500 mol.
m(H₂O) = 0,5 mol · 18 g/mol.
m(H₂O) = 9 g.
3) m(total) = 8,24 g + 9 g + 7,23 g.
m(total) = 24,47 g.
n - amount of substance.
Answer:
The answer to your question is C₃H₆O
Explanation:
Data
mass of sample = 23.2 g
mass of carbon dioxide = 52.8 g
mass of water = 21.6 g
empirical formula = ?
Process
1.- Calculate the mass and moles of carbon
44 g of CO₂ --------------- 12 g of C
52.8 g --------------- x
x = (52.8 x 12)/44
x = 633.6/44
x = 14.4 g of C
12 g of C ------------------ 1 mol
14.4 g of C --------------- x
x = (14.4 x 1)/(12)
x = 1.2 moles of C
2.- Calculate the grams and moles of Hydrogen
18 g of H₂O --------------- 2 g of H
21.6 g of H₂O ------------- x
x = (21.6 x 2) / 18
x = 2.4 g of H
1 g of H -------------------- 1 mol of H
2.4 g of H ----------------- x
x = (2.4 x 1)/1
x = 2.4 moles of H
3.- Calculate the grams and moles of Oxygen
Mass of Oxygen = 23.2 - 14.4 - 2.4
= 6.4 g
16 g of O ---------------- 1 mol
6.4 g of O -------------- x
x = (6.4 x 1)/16
x = 0.4 moles of Oxygen
4.- Divide by the lowest number of moles
Carbon = 1.2 / 0.4 = 3
Hydrogen = 2.4/ 0.4 = 6
Oxygen = 0.4 / 0.4 = 1
5.- Write the empirical formula
C₃H₆O
Answer:
- 0.0249% Sb/cm
Explanation:
Given that:
One surface contains 1 Sb atom per 10⁸ Si atoms and the other surface contains 500 Sb atoms per 10⁸ Si atoms.
The concentration gradient in atomic percent (%) Sb per cm can be calculated as follows:
The difference in concentration = 
The distance 
= 0.2-mm = 0.02 cm
Now, the concentration of silicon at one surface containing 1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per 10⁸ silicon atoms can be calculated as follows:
= - 0.0249% Sb/cm
b) The concentration 
of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:
Lattice parameter = 5.4307 Å; To cm ; we have
= 
= 
The concentration 
of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:
= 
= 
Finally, to calculate the concentration gradient
Answer:
Explanation:
wavelength λ = 12.4 x 10⁻² m .
energy of one photon = h c / λ
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 12.4 x 10⁻²
= 1.6 x 10⁻²⁴ J .
Let density of coffee be equal to density of water .
mass of coffee = 255 x 1 = 255 g
heat required to heat up coffee = mass x specific heat x rise in temp
= 255 x 4.18 x ( 62-25 )
= 39438.3 J .
No of photons required = heat energy required / energy of one photon
= 39438.3 / 1.6 x 10⁻²⁴
= 24649 x 10²⁴
= 24.65 x 10²⁷ .
