3
One from fe
And two from cl2
Answer:
I think Sammy is right because I'm sure the compound contains ionic bonds
The number of H atoms in 3(NH₄)₂CrO₄ = 24
<h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
Subscripts in the chemical formula indicate the number of atoms
The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :
Number of H :
![\tt 4\times 2(subscript)\times 3(coefficient,number~of~molecules)=24~atoms](https://tex.z-dn.net/?f=%5Ctt%204%5Ctimes%202%28subscript%29%5Ctimes%203%28coefficient%2Cnumber~of~molecules%29%3D24~atoms)
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>