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
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Answer:
9.54
Explanation:
Fr the question given above, the following data were:
Concentration of hydronium ion [H3O⁺] = 2.9x10¯¹⁰ M
pH =?
The pH of the solution can be obtained as follow:
pH = –Log [H3O⁺]
pH = –Log 2.9x10¯¹⁰
pH = 9.54
Thus, the pH level of the solution is 9.54
<span>Mr = 13 g / mol
mass = 5 g
Mol = 5/13 mol :)</span>
Answer: The answer is True
Explanation: I hope my answer helps :)
Answer:
We are given:
Volume (V) = 0.25 L
Pressure (P) = 0.93 atm
Temperature (T) = 15.4°C OR 288.4 K
<u>Solving for the number of moles of CO₂:</u>
From the ideal gas equation:
PV = nRT
replacing the variables
0.93 * 0.25 = n (0.082)(288.4)
n = 0.00983 moles
<u>Number of molecules:</u>
Number of moles= 0.00983
number of molecules in 1 mole = 6.022 * 10²³
Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³
Number of molecules = 5.91 * 10²¹
