The two magnets ought to join together given that at least, if not both, have a strong enough magnetic force to do so.
The amount of sugar is 2621 mg
Why?
The complete question is:
A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.
Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?
To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:
Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:
Have a nice day!
Answer:
A) SiO2 is the limiting reactant
B) Theoretical yield= 72333.3g
C) % yield =91.5%
Explanation:
SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)
n(SiO2)= 155000/60 = 2583.33 mols
n(C)= 79000/12= 3291.66 mols
a)SiO2 is the limiting reactant
According to the balanced reaction equation,
60g of SiO2 produced 28g of SiO2
155000g of SiO2 will produce 155000×28/60= 72333.3g
Therefore theoretical yield of Si= 72333.3g
% yield= 66200/72333.3×100/1 =91.5%
For plato users
the answer is a. O2(l) O2(g)
hope this helps!
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
