step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%
step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2
To calculate how many photons are in a certain amount of energy (joules) we need to know how much energy is in one photon.
Start by using two equations:
Energy of a photon = Frequency * Planck's constant (6.626 * 10^(-34) J-s)
Speed of light (constant 3 * 10^8 m/s) = Frequency * Wavelength
Which means:
frequency = Speed of Light / Wavelength
So energy of a photon = (Speed of light * Planck's constant)/(Wavelength)
You may have seen this equation as E = hc/<span>λ</span>
We have a wavelength of 691 nm or 691 * 10^-9 meters
So we can plug in all of our knowns:
E = (6.626 * 10^(-34) J-s) * (3.00 * 10^8 m/s) / (691 * 10^-9 m) =
2.88 * 10^(-19) joules per photon
Now we have joules per photon, and the total number of joules (0.862 joules)
,so divide joules by joules per photon, and we have the number of photons:
0.862 J/ (2.88 * 10^(-19) J/photon) = 3.00 * 10^18 photons.
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
