Answer:
V2 = 2.88L
Explanation:
P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K
Applying the general gas equation
P1V1/T1 = P2V2/T2
Substitute the above
(78*2)/900= (45*V2)/750
V2= (78*2×750)/(900*45)
V2= 2.88L
1.1kg = 1100g
If each paper clip is 0.88g : 1100 / 0.88 = 1250
1250 paper clips in the box
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
<span>D. The average kinetic energy of their particles is the same.</span>
