Answer:
1 mole of iron =6.023×10^23 particles
1 particles of iron=1/6.023×10^23 mole
7.46×10^25 particles =1/6.023×10^23×7.46×10^25
=1.238×10^48 mole is a required answer.
This is valid only if the two elements have the same valence
Data Given:
Initial Volume = V₁ = 36.7 L
Initial Pressure = P₁ = 145 kPa
Initial Temperature = T₁ = 65 °C + 273 = 338 K
Final Volume = V₂ = ?
Final Pressure = P₂ = 101.325 kPa (Standard Pressure)
Final Temperature = T₂ = 273 K (Standard Temperature)
Formula used:
As number of moles are constant, so Ideal Gas equation in following form is used,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / T₁ P₂
Putting Values,
V₂ = (145 kPa × 36.7 L × 338 K) ÷ (273 K × 101.325 kPa)
V₂ = 1798667 ÷ 27661.25
V₂ = 65.02 L
Answer:
Explanation:
98 gH
2
SO
4
contains1 mole of H
2
SO
4
.
1 mole of H
2
SO
4
contain=1 mole of sulphur atoms.
∴9.8 g of H
2
SO
4
contain =0.1 mol of sulphur atom.
Thus, number of sulphur-atoms =n×N
A
(in 9.8 g of H
2
SO
4
)
=0.1×6.023×10
23
=0.6023×10
23
.
