A single sodium ion will combine due to its charge on it
Na^+
The single sodium ion has total one positive charge on it. so it requires only one electron or can bond to one negative charge
in the given polyatomic anions
a) CO3^-2 : it has two negative charge. So it will react with two sodium ions
it will forms Na2CO3
b) PO4^-3 :it has three negative charge. So it will react with three sodium ions
it will forms Na3PO4
c) SO4^-2 : it has two negative charge. So it will react with two sodium ions
it will forms Na2SO4
d) NO3^-1: it has one negative charge. So it will react with one sodium ion
it will forms NaNO3
Hence the correct answer is
Nitrate ion will react with single sodium ion as
Na+ + NO3- ---> NaNO3
Explanation:
The given data is as follows.
Concentration of standard NaOH solution = 0.1922 M
Let the original acid solution concentration be x.
= 0.1 x M
= 10.00 mL (given)
The reaction equation is as follows.
Concentration × Volume of 
= Concentration × Volume of NaOH
x = 1.314 M
Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.
Answer:
I didn't understand what u said
1) All the CO2 comes from the C24 H30 N3 O
2) To balance C from CO2 with C from C24, the ratio of cC24 H30 N3 O to CO2 is 24 / 1
3) Convert 1.20 g of CO2 to number of moles
number of moles = mass in grams / molar mass
mass = 1.20 g
molar mass = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
number of moles = 1.2 g / 44.01 g/mol = 0.0273 mol
4) proportion
1 mol C24 H30 N3 O / 24 mol CO2 = x / 0.0273 mol CO2
=> x = 0.0273 mol CO2 * 1 mol C24H30N3O / 24 mol CO2 = 0.00114 mol C24H30N3O
5) Convert 0.00114 mol C24H30N3O to grams
molar mass = 376.5 g/mol
mass = 0.00114 mol * 376.5 g/mol = 0.429 g
6) mass percent in the mixture
mass percent = (mass of lsd / mass of mixture) * 100 = (0.429g / 1.0g) * 100 = 42.9%
Answer: 42.9%
