Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.
Normality comes out to be 8.11
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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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I believe the answer is an ion.
Would it be standard notation
Answer:
1.09 grams
Explanation:
According to the following chemical equation:
HF + NaNO₃ -> HNO₃ + NaF
1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:
MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF
1 mol HF x 19.9 g/mol HF = 20 g
MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF
1 mol NaF x 42 g/mol NaF = 42 g
Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:
20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF
Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.
