I believe it’s Refractive index
Answer is: mass of butane is D)11.6 g.
m(butane) = 50,0 g.
V(CO₂) = 17,9 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
From chemical reaction n(CO₂) : n(C₄H₁₀) = 8 : 2.
n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
m(C₄H₁₀) = 0,2 mol · 58 g/mol.
m(C₄H₁₀) = 11,6 g.
Answer:
yes
Explanation:
the results are CO2 and H2O, which usually appear after some kind of combustion.(combustion=burning)
hope this helps!
