3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>
Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :

Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³