Answer:
The answer to your question is empirical formula Al₃O₉S
Explanation:
Data
Al = 31.5 %
O = 56.1 %
S = 12.4 %
Process
1.- Look for the atomic masses of the elements
Al = 27 g
O = 16
S = 32
2.- Represent the percentages as grams
Al = 31.5 g
O = 56.1 g
S = 12.4 g
3.- Convert these masses to moles
27 g of Al ----------------- 1 mol
31.5 g ---------------------- x
x = 1.17 moles
16 g of O ---------------- 1 mol
56.1 g of O ------------- x
x = 3.5 mol
32 g of S --------------- 1 mol
12.4 g of S ------------- x
x = 0.39 moles
4.- Divide by the lowest number of moles
Al = 1.17 / 0.39 = 3
O = 3.5 / 0.39 = 8.9 ≈ 9
S = 0.39 / 0.39 = 1
5.- Write the empirical equation
Al₃O₉S
Answer:
a. 10.54
Explanation:
reaction is
NH₃ + NaOH -----------------NH₄Cl + H₂O
0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl
mol NH₃ = 0.80 - 010 = .70
mol NH₄Cl = 0.80 + .10 = 0.90
pH = pKₐ + log (( NH₃/NH₄Cl))
pH = 9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54
<span>You could find the volume of an irregular object by water displacement. You measure the initial volume of water in a beaker, then you place the object in the water. Record the final volume and subtract it from the initial, and there you have the volume of the object.</span>
CI2 + 2Nal ---> 2NalCI + I2
Nal = 149.9 g/mol.
NaCI = 58.5 g/mol
2 mol Nal makes 2 mol NaCI
0.29 g Nal x ( 1 mol / 149.9 g) = 0.0019 mol NaCI
00.19 mol NaCI x ( 58.5 g / 1 mol ) 0.11 g NaCI ( 2 sig figs)
