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
The easiest way to answer this question is to first figure out the molar mass of the sugar in question. To do this multiply the number of individual atoms for a given element by its atomic mass. 12 X 12.01 g/mol = 144.12 g/mol C. 22 X 1.008 g/mol = 22.176 g/mol H. 11 X 16.00 g/mol = 176.00 g/mol O.
The best answer is b) yess it’s b try it it’s the best one I’m pretty sure
Answer:
V₂ = 72.2 mL
Explanation:
Given data:
Initial volume = 146.0 mL
Initial pressure = 1.30 atm
Initial temperature = 5.00 °C (5.00 +273 = 278 K)
Final temperature = 2.00 °C ( 2.00+273 = 275 K)
Final volume = ?
Final pressure = 2.60 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 1.30 atm × 146.0 mL × 275 K / 278K × 2.60 atm
V₂ = 52195 atm .mL. K / 722.8 K.atm
V₂ = 72.2 mL