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>
A. immigration, how is this chemistry?
Answer:
3
Explanation:
Applying,
= R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
= 40/5
= 8
= 2³
Equation the base,
n' = 3
The balanced equation will read:
Pb(NO3)2 (aq) + LiSO4 (aq) ➡️ PbSO4 (s) + 2️⃣LiNO3 (aq)
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