Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
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:
Pb(NO3)2
Cd(NO3)2
Na2SO4
Explanation:
In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.
When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.
After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.
The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.
Answer: Thus 24.0 g of
would be needed.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of
require = 2 moles of
Thus 0.1875 moles of
will require=
of
Mass of
Thus 24.0 g of
would be needed to completely react with 6.00 g of
such that all reactants could be consumed.