Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
<em>Moles C6H12O6:</em>
650g * (1mol/180.16g) = 3.608 moles C6H12O6
<em>Moles O2:</em>
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
<h3>O2 is limiting reactant</h3>
Answer:
54.30 grams acetone
Explanation:
(0. 935 mol)*(58.08 g/mole) = 54.30 grams acetone
Independent would be the amount of sugar given and the dependent would be the amount of cavities
Scientists have control groups so they can have evidence to show the difference between that one and the experimental ones.
And the independent variable is the one that changes because with out it changing you wouldn't get different results.
I could be wrong tho sorryyy! but i hope this helps
Answer:
glucose and fructose
Explanation:
Sucrose is a disaccharide (a kind of sugar made of two monosaccharides) made of glucose and fructose
M(P)=3.72 g
M(P)=31 g/mol
m(Cl)=21.28 g
M(Cl)=35.5 g/mol
n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol
n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol
P : Cl = 0.12 : 0.60 = 1 : 5
PCl₅ - is the empirical formula
