Answer:
At anode - 
At cathode - 
Explanation:
Electrolysis of NaBr:
Water will exist as:
The salt, NaBr will dissociate as:
At the anode, oxidation takes place, as shown below.
At the cathode, reduction takes place, as shown below.
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
Na2C03 is formed through the reaction of
NaOH and H2CO3 namely sodium hydroxide and carbonic acid
NaOH -> strong base
H2CO3-> weak acid
To solve this kinematics formula use the following equation:
Vf = Vi + at
Vf = 0 + (9.81 m/s^2)(3 seconds)
Vf = 29.43 m/s and or about 29.4 m/s of reported to 3 significant figures.
Answer:
50?
I think but not 100% sure if not lmk
