Use formula n= number of particles / avogrado constant
n= number of mole
avogrado constant= 6.12x10^-23
you will get 2.042 x 10^52
Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.
Normality comes out to be 8.11
<h3>
Given </h3>
- Mass of solute: 1000g
- Volume of solution (V): 5000 ml = 5 liters
- Equivalent mass of solute (E) = molar mass / n-factor
n-factor for 
is 6 and molar mass is 148g
So, on calculating equivalent mass is equal to 24.66g
FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)
N=
<u> N=8.11</u>
Therefore, normality of 1 kg aluminum sulfide is 8.11
Learn more about normality here brainly.com/question/25507216
#SPJ10
It does that because petrol is a liquid chemical,has a lower mass per volume,or density,then the water on top so it spread out on top of the water
<h2>Answer:</h2><h3>Part 1. </h3>
Option B is correct option.
The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing OH- ions.
Explanation:
Full equation:
2MnO2 + H2O + 2e- → Mn2O3 + 2OH-
<h3>Part 2:</h3>
The option B which is Mg is stronger reducing agent than Ag is correct option.
Explanation:
Equation:
Mg(s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag(s)
According to equation Mg converts to Mg+2 which means it gives to electron to reduce Ag. So it act as an reducing agent.
<h3>Part 3:</h3>
The correct option is B. Which is 5, 1, 8, 5, 1, 4.
Explanation:
Full equation :
5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) --> 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O (l)
<h3 />
