Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is 
Answer:
Ka = 
Explanation:
Initial concentration of weak acid = 
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:
(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
0.000000135 <<< 0.00045
I will list them from alkaline with the lowest boiling point and alkaline with the highest.
1. C2H6
2. C9H20
3. C11H24
4. C16H34
5. C20H42
6. C32H66
7. C150H302
I have taken a quiz similar to this before and can assure you this is correct and is primarily because of the number of Carbons and Hydrogens within this. More Carbons and Hydrogens causes Boiling Points to increase because of stronger bonds.
the correct scientist is rutherford
First write the balanced equation of this reaction:
2H2 + O2 —> 2H2O
mol of H2= 0.60 gH2/2.02 gH2 = 0.297 mol
There are 2 mol of H2 for every 2 mol of H2O so the number of mol of H2 is equal to the number of mol of H2O.
g of H2O = 0.297 mol H2O • 18.02 gH2O = 5.35 g H2O
Do the same thing for O2:
mol of O2 = 4.8 gO2/32.0 gO2 = 0.15 mol of O2
There is 1 mol of O2 for every 2 mol of H2O so multiply 0.15 • 2 to get the number of mol of H2O
g of H2O = 0.30 mol H2O • 18.02 gH2O = 5.41 g H2O
The correct answer is 5.35 g H2O (or 5.4 g if checking significant figures) because O2, in this case, is the limiting reactant of this reaction.
