Answer:
Ammonia, urea, uric acid
Explanation:
The given compounds are:-
Urea which has a molecular formula of
and has 2 atoms of nitrogen per molecule.
Ammonia which has a molecular formula of
and has 1 atom of nitrogen per molecule.
Uric acid which has a molecular formula of
and has 4 atoms of nitrogen per molecule.
Thus,
The order from least to most nitrogen is:-
Ammonia, urea, uric acid
Answer:
11
Explanation:
Moles of KOH = 
Volume of water = 10 liters
Concentration of KOH is given by
![[KOH]=\dfrac{10^{-2}}{10}\\\Rightarrow [KOH]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5Cdfrac%7B10%5E%7B-2%7D%7D%7B10%7D%5C%5C%5CRightarrow%20%5BKOH%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
is strong base so we have the following relation
![[KOH]=[OH^{-}]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E%7B-%7D%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
![pOH=-\log [OH^{-}]=-\log10^{-3}](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E%7B-%7D%5D%3D-%5Clog10%5E%7B-3%7D)

So, pH of the solution is 11
The reaction is,
H2S + I2 --------------> 2 HI +S
Molar weight of H2S = 34 g per mol
Molar weight of HI =128 g per mol
Molar weight of I2 =254 g per mol
Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol
So according to stoichiometry of the reaction, number of I2 mols needed
= 1.447 mol
The mass of I2 needed = 1.447 mol x 254 g