Answer:
Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula
Explanation:
This is the reaction:
NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O
In 100 g of compound we have:
28.2 g N
20.8 g of P
8.1 g of H
42.9 g of O
Now we divide each between the molar mass:
28.2 g / 14 g/mol = 2.01 mol
20.8 g / 30.97 g/mol = 0.671 mol
8.1 g / 1 g/mol = 8.1 mol
42.9 g / 16 g/mol = 2.68 mol
And we divide again between the lowest value of moles
2.01 mol / 0.671 mol → 3
0.671 mol / 0.671 mol → 1
8.1 mol / 0.671 mol → 12
2.68 mol / 0.671 mol → 4
Molecular formula will be: N₃PH₁₂O₄ → (NH₄)₃PO₄
Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula
its the oceam which is blue in color so
Answer:
HCl conc.= 6.0mol/L
Explanation:
From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,
Answer:
2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)
Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?
ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol
Answer:
0.01 M
Explanation:
The chemist is performing a serial dilution in order tyo obtain the calibration curve for the instrument.
First we must obtain the concentration of the solution in the 250ml flask from
C1V1 = C2V2
Where;
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the diluted solution
V2= volume of the diluted solution
2.61 × 10 = C2 × 250
C2 = 2.61 × 10/250
C2 = 0.1 M
Hence for solution in 100ml flask;
0.1 × 10 = C2 × 100
C2 = 0.1 × 10/100
C2 = 0.01 M
