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Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=
Answer:
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is :
concentration of = 10%
10 g of in 100 ml solution
molar mass = 45.08 g/mol
number of moles = 10 / 45.08
= 0.222 mol
Molarity of
= 2.22 M
number of moles of in 20 mL can be determined as:
Concentration of
= 2.22 M
Similarly, The pKa Value of is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
the pH value at equivalence point is,
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6