step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%
step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
Thank you for posting your question here brainly. Based on the problem mentioned above the largest mass that water molecule could have using other isotopes is <span>24 amu. Below is the solution, I hope the answers helps.
</span><span>T2_18O = 24</span>
The given question is incomplete. The complete question is:
A chemist prepares a solution of barium chloride by measuring out 110 g of barium chloride into a 440 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mole per liter of the chemist's barium chloride solution. Round your answer to 3 significant digits.
Answer: Concentration of the chemist's barium chloride solution is 1.20 mol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
moles of 
(solute) = 
Now put all the given values in the formula of molality, we get
Therefore, the molarity of solution is 1.20 mol/L
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
