B gets multiplied 3 times as much aswell. (brainly pls) :D
Molar mass NaCl = 58.44 g/mol
number of moles:
mass NaCl / molar mass
145 / 58.44 => 2.4811 moles of NaCl
Volume = 3.45 L
Therefore :
M = moles / volume in liters:
M = 2.4811 / 3.45
M = 0.719 mol/L⁻¹
hope this helps!
Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
56.4 mmHg
Explanation:
Given:
Vapor pressure of the solution, P solution = 55 mmHg
The mass of sucrose (C₆H₁₂O₆) = 10 g
Also, Molar mass of sucrose (C₆H₁₂O₆) = 180 g/mol
So, moles = Given mass/ molar mass
Hence, moles of sucrose in the solution = 10 g / 180 g/mol = 0.05556 mol
Given that: Mass of ethanol = 100 g
Molar mass of ethanol = 46 g/mol
Hence, moles of ethanol = 100 g / 46 g/mol = 2.174 mol
Mole fraction of solvent, ethanol is:
X ethanol = 2.174 mol / (2.174 + 0.05556) mol = 0.975
Applying Raoult's Law
P solution = X ethanol*P° ethanol
<u>
=> P° ethanol = P solution / X ethanol = 55 mmHg / 0.975 = 56.4 mm Hg</u>
