1 gram of HNO3 will be equal to 1/63.01 moles. Therefore, we can say that 1 liter of Nitric acid contains 15.6976 moles or in other words molarity of 70% (w/w) Nitric acid is equal to 15.6976 M.
Known values
Molecular weight of HNO3 63.01 g/mole
Concentration of Nitric acid 70% (% by mass, wt/wt)
Question options:
A) K2SO4
B) FeCl₃
C) NaOH
D) NH₃
E) KCl
Answer:
D. NH₃
Explanation:
K2SO4 = 2 K+ + SO42-
[K+]= 2 x 1.0 = 2.0 M ; [SO42-] = 1.0 M
total concentrations of ions = 2.0 + 1.0 = <em>3.0 M</em>
FeCl3 = Fe3+ + 3Cl-
[Fe3+] = 1.0 M ; [Cl-] = 3 x 1.0 = 3.0
total concentration ions = 1.0 + 3.0 =<em> 4.0 M</em>
NaOH = Na+ + OH-
[Na+] = [OH-] = 1.0 M
total concentration ions = 1.0 + 1.0 = <em>2.0 M</em>
<u>NH3 is a weak acid so the concentration of NH4+ and OH- </u><u><em>< 2.0</em></u>
KCl = K+ + Cl-
[K+] = [Cl-] = 1.0 M
total concentration ions = 1.0 + 1.0 =<em> 2.0 M</em>
Because they do not have the same qualities therefore they are different
The balanced chemical reaction is written as :
Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:
120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
Protein would be the main one.
