Answer:
b
Explanation:
Atom is the smallest particle of an element
0.025 is how many moles of oxygen to react with NH3
Answer:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2
Explanation:
Ca3(PO4)2 + H2SO4 —> CaSO4 + H3PO4
From the above equation,
There are 3 atoms of Ca on the left and 1 atom of Ca on the right. To balance Ca, put 3 in front of CaSO4 as shown below
Ca3(PO4)2 + H2SO4 —> 3CaSO4 + H3PO4
Now, we have 3 atoms of SO4 on the right and 1atom on the left. To balance SO4, put 3 in front of H2SO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + H3PO4
Looking closely, there are 6 atoms of H on the left and 3 on the right. Therefore, it is balanced by by putting 2 in front of H3PO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2
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
