The result will be 100.60 because we are adding to the atmosphere, which connects to the lubricator, and then it turns, resulting in 100.60.
0.1045M represents the concentration of the initial sulfuric acid solution
Explanation:
Titration is conducted to ascertain the volume or concentration of an unknown electrolyte.
Given data:
acid volume = 50 ml
acid molarity =?
NaOH volume = 34.62 ml
molarity of NaOH = 0.1510
The titration formula applied is
Macid x Vacid = Mbase x Vbase
Substituting the values into the above equation results in:
Macid x 50 = 34.62 x 0.1510
Macid = 
= 0.1045 M, indicating the molarity of the sulfuric acid solution used to neutralize the 0.1510 M base solution.
Thanks for the responses ッ. (By the way, the answers are located at the bottom of the question if anyone hasn’t noticed them.
The percentage of KCl present in the mixture is approximately <span>40%</span><span>If we consider a 100-gram sample of the mixture:ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.</span>n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equations:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From the first equation, we find x = 100 - y and substitute into the second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
Thus, x = m(KCl) = 100 g - 62.83 g = 37.17 g.
1) Chromium(III) nitrate is classified as acidic because it is derived from a weak base (chromium(III) hydroxide Cr(OH)₃) and a strong acid (nitric acid HNO₃). 2) Sodium hydrosulfide is considered basic because it results from a strong base (sodium hydroxide NaOH) and a weak acid (hydrogen sulfide H₂S). 3) Zinc acetate is slightly basic since zinc hydroxide (Zn(OH)₂) is a stronger base than acetic acid (CH₃COOH).
