Hello! The correct answer is, B. are in motion outside the nucleus.
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Answer:
=1.666 liters
Explanation:
1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.
0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1
=11.2 liters.
Standard pressure= 1 atmosphere (Atm)
Standard temperature = 273.15 Kelvin
According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂
Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.
Therefore P₂ =P₁V₁T₂/T₁V₂
Substituting for the values we get:
P₂= (1 atm× 11.2L ×203K)/ (273K×5L)
=1.666 atm
The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams