Answer:
pH of the final solution = 9.15
Explanation:
Equation of the reaction: HCl + NH₃ ----> NH₄Cl
Number of moles of NH₃ = molarity * volume (L)
= 0.4 M * (300/1000) * 1 L = 0.12 moles
Number of moles of HCl = molarity * volume (L)
= 0.3 M * (175/1000) * 1 L = 0.0525 moles
Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced
Number moles of NH₄Cl produced = 0.0525 moles
Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675
pOH = pKb + log([salt]/[base])
pKb = -logKb
pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)
pOh = 4.744 + 0.109
pOH = 4.853
pH = 14 - pOH
pH = 14 - 4.853
pH = 9.15
Therefore, pH of the final solution = 9.15
Answer:
93.33 g of Fe
Explanation:
The balanced equation for the reaction is given below:
4Fe + 3O₂ —> 2Fe₂O₃
Next, we shall determine the masses of Fe and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of Fe = 56 g/mol
Mass of Fe from the balanced equation = 4 × 56 = 224 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 3 × 32 = 96 g
SUMMARY:
From the balanced equation above,
224 g of Fe reacted with 96 g of O₂.
Finally, we shall determine the mass of Fe required to react with 40 g of O₂. This can be obtained as follow:
From the balanced equation above,
224 g of Fe reacted with 96 g of O₂.
Therefore, Xg of Fe will react with 40 g of O₂ i.e
Xg of Fe = (224 × 40)/96
Xg of Fe = 93.33 g
Therefore, 93.33 g of Fe is required to react with 40 g of O₂.
Answer:
E= h*frequency
6.626*10^-34)*5.71*10^14)= 3.78*10^-19 J
Answer:
Explanation:
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