Convert each amount of grams into moles:
I: 23.24g x 1 mol / 126.90g = 0.1831 mol I
C: 2.198 x 1 mol / 12.01g = 0.1830 mol C
N: 2.562 x 1 mol / 14.01g = 0.1829 mol N
Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1
Therefore the empirical formula is ICN
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
Answer: the HCO3- to act as a base and remove excess H by the formation of H2CO3
Explanation:
H2CO3 in an aqueous solution is a buffer. This means the reaction is the following:
H2CO3 ------ HCO3- + H+
Then, the HCO3- that was formed acts as a base (absorbing a proton) like this
HCO3- + H+ ------- H2CO3
If there was an increase in H+, there would be an increase in the second reaction in an effort to neutralize that acid, thus making the H2CO3 more concentrated
Answer: 3.36 L of ammonia gas
Explanation:
The balanced chemical reaction is:
According to stoichiometry :
3 moles of 
produce = 2 moles of 
Thus 0.75 moles of will producee=
of
But as percent yield is 30 %, amount of ammonia produced = 
According to ideal gas equation:
P = pressure = 1 atm
V = Volume = ?
n = number of moles = 0.15
R = gas constant =
T =temperature =
Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.
