Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
= 207+56+192 = 455 g/mol
Formula:
Solving:
Answer:
<span>
B. 0.119 M</span>
I’m assuming your just writing the formula? If so
Potassium chloride: KCL
Potassium nitride: KNO2
Potassium sulfide: K2S
calcium chloride: CaCl2
Calcium nitride: Ca3N2
Calcium sulfide: CaS
Silver chloride: AgCl
Silver nitride: Ag3N
Silver sulfide: Ag2S
Manganese (||) chloride: MnCl2
Manganese (||) nitride: Mn3N2
Manganese (||) sulfide: MnS
Answer:
option C= 12.40
Explanation:
Formula:
pH + pOH = 14
First of all we will calculate the pH.
pH = - log [H⁺]
pH = - log [0.025]
pH = - (-1.6)
pH = 1.6
Now we will put the values in formula,
pH + pOH = 14
pOH = 14-pH
pOH = 14 -1.6
pOH = 12.4
The pOH of solution is 12.4.
Main group elements
transition metals
lanthanides
actindes.