How is the number 8.697 x 10-5 expressed in regular numerals? C. 0.08697
Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.
From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
= mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
=
= mol
Mass of ethanol
=
=
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
= mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed= mol
Mass of ethanol formed
=
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
Measured values are verified and go under a series of experiments before being accepted as the norm. Accepted values are unanimously accepted values, seen as the norm by a qualified group of individuals (scientists or the people of the intellectual sphere).
Answer:
0.05 mL
Explanation:
Initially, we have a concentrated solution of NaOH, to which we will add water to get a dilute one. To calculate the volume of the initial solution that we have to measure, we can use the dilution rule:
C₁ × V₁ = C₂ × V₂
where,
C₁ is the initial concentration (50% w/w; ρ = 1.52 g/mL)
V₁ is the initial volume
C₂ is the final concentration (0.1 M)
V₂ is the final volume (10 mL)
First, we have to calculate the weight/volume percentage and then the molarity of the first solution.
Now, we can apply the dilution rule.