Spent fuel that can no longer be used to create energy is waste.
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
Explanation:
A low-pressure area, or "low", is a region where the atmospheric pressure at sea level is below that of surrounding locations. Low-pressure systems form under areas of wind divergence that occur in upper levels of the troposphere.
<span>Now consider a low pressure area on a disk as shown below.A parcel of air at point A would move toward the center of the low pressure area. That movement would take it farther away from the center of the disk and therefore it would move to the west. A parcel of air at B would move toward the center of the low pressure area which would also take it closer to the center of the spinning disk where its speed is greater than the surrounding points. It would appear to move to the east. With A moving to the west and B moving to the east the line from A to B is rotating counterclockwise.</span>
Answer: Salt and Water
Explanation:
An Arrhenius acid (HCl) can best be defined as any substance that when added to water increases the concentration of H+ ions.
While an Arrhenius base (KOH) is any substance that when added to water increases the concentration of OH- ions.
When an Arrhenius acid such as HCl reacts with an Arrhenius base such as KOH, the end products will be salt and water, in a process called Neutralization Reaction.
HCl (aq) + KOH (aq) -------> KCl (aq) + H2O (l)
