D. The number increases and then decreases for noble gases
Answer:
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Explanation:
Moles of ethanol of ethanol = 0.200 mol
Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ
Enthalpy of vaporization of ethanol = 
Heat required to vaporize 0.200 moles of ethanol = Q'
Total heat required to fore heating and the vaporizing :
= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Answer:
cW = 4.2J/g.k
Explanation:
Heat gained = Heat lost
iron(mc∆€) = water(mc∆€)
mi= 100g
ci = 0.449
∆€i= 150-25 = 125
...
mw= 268.5g
cw = ?
∆€w=25-20 = 5
...
100×0.449×125= 268.5×cw×5
5612.5= 1342.5cw
cw = 5612.5/1342.5
cw= 4.18 ~ 4.2J/g.K
Answer:
Ksp = 2.4 * 10^-13
Explanation:
Step 1: Data given
Molarity of NaIO3 = 0.10 M
The molar solubility of Pb(IO3)2 = 2.4 * 10^-11 mol/L
Step 2: The initial concentration
NaIO3 = 0.1M
Na+ = 0 M
2IO3- = 0 M
Step 3: The concentration at the equilibrium
All of the NaIO3 will react (0.1M)
At the equilibrium the concentration of NaIO3 = 0 M
The mol ratio is 1:1:1
The concentration of Na+ and IO3- is 0.1 M
Pb(IO3)2 → Pb^2+ + 2IO3^-
The concentration of Pb(IO3)2 can be written as X
The concentration of Pb^2+ can be written as X
The concentration of 2IO3^- can be written as 2X
Ksp = (Pb^2+)(IO3^-)²
⇒ with (Pb^2+) = 2.4*10^-11
⇒ with (IO3^-) = 2x from the Pb(IO3)2 and 0.1M from the NaIO3.
⇒The total (IO3^-) = 2x + 0.1 and we assume that x is so small that we can neglect it.
Ksp = (2.4 *10^-11)*(0.1)²
Ksp = 2.4 * 10^-13
Answer:
For mole conversion always remember 1 mole of compound = molar mass of compound.
Ex. 2 moles H20 to grams
2mol H2O x 18 g / 1mol H20
For molecules use Avogadro's number like moles multiplied to 6.022x10²³ answer will be in molecules.
Explanation:
