The output density is given as kg/m 3, lb/ft 3, lb/gal(US liq) and sl/ft 3. Specific weight is given as N/m 3 and lb f / ft 3.
A stoichiometric mixture is a mixture of fuel and oxygen for which the masses of these two components are exactly those needed for complete combustion.
A stoichiometric mixture is a balanced mixture of fuel and oxygen.
The fuel and the oxygen react completely without the excesses of either.
The opposite of a stoichiometric mixture is called feeding an excess, when minimum one reactant is an excess amount.
Balanced chemical equation for reaction of combustion one type of a fuel: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
Stoichiometric mixture for this example is when fuel (C₈H₁₈) and oxygen(O₂) react in proportion 1 : 12.5.
More about a stoichiometric mixture: brainly.com/question/19585982
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C. dim light is <span>the answer
</span>
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.
the correct answer is b it makes the most sense
