Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
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Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of butane

b) moles of oxygen


According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of 
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
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