Answer:
The last region should be right
Explanation:
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>
Answer:
the answer to the qustion is 0.013089701 na
Explanation:
n/a
Answer:
Proteins and nucleic acids
Explanation:
Nitrogen compounds in animals that are no longer of use, or are in access are excreted from the animals body, and are thus called nitrogenous waste. These nitrogenous waste can be excreted in three different ways.
1. Ammonia
2. Urea
3. Uric acid