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(NO_3)_4](https://tex.z-dn.net/?f=Pb%28NO_3%29_4)
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
![Pb(NO_3)_4](https://tex.z-dn.net/?f=Pb%28NO_3%29_4)
= 207+56+192 = 455 g/mol
Formula:
![M = \frac{m}{MM*V}](https://tex.z-dn.net/?f=M%20%3D%20%20%5Cfrac%7Bm%7D%7BMM%2AV%7D%20)
Solving:
![M = \frac{m}{MM*V}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bm%7D%7BMM%2AV%7D%20)
![M = \frac{13.50}{455*0.25}](https://tex.z-dn.net/?f=M%20%3D%20%20%5Cfrac%7B13.50%7D%7B455%2A0.25%7D%20)
![M = \frac{13.50}{113.75}](https://tex.z-dn.net/?f=M%20%3D%20%20%5Cfrac%7B13.50%7D%7B113.75%7D%20)
![M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=M%20%3D%200.118681318...%5C%3A%5C%3A%5Cto%5C%3A%5C%3A%5Cboxed%7B%5Cboxed%7BM%20%5Capprox%200.119%5C%3AMol%2FL%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Answer:
<span>
B. 0.119 M</span>
O,P,Ge ranked from atomic radius
Erosion
waethring
deposition
pressure or heat
all of this can change the form of rock to another
Answer:
Round to the number of significant figures in the original question. However, if you're going to proceed with further calculations using this mass, it's best not to round, as rounding will cause your answer to be less precise.
Explanation: