i love math but sadly i am dumb
Answer:
30.0 mol CO₂
Explanation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
To answer this problem we need to convert moles of C₃H₈ into moles of CO₂: We'll do that by using the <u>stoichiometric coefficients</u>, using a conversion factor that has C₃H₈ moles in the denominator and CO₂ moles in the numerator:
10.0 mol C₃H₈ * 
= 30.0 mol CO₂
Think it is b
but i am not sure at all
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
