The question is incomplete, here is the complete question:
Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in 
The image is attached below.
<u>Answer:</u> The density of CsI is 
<u>Explanation:</u>
To calculate the density of metal, we use the equation:
where,
= density
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of CsI = 259.8 g/mol
= Avogadro's number = 
a = edge length of unit cell = 
(Conversion factor: 
)
Putting values in above equation, we get:
Hence, the density of CsI is
Answer:
1.20 V
Explanation:
The standard cell potential is calculated from the expression
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction
The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.
Thus for our question we will have
oxidation:
Pb(s) → Pb2+(aq) + 2 e- ε⁰ oxidation = - ε⁰ reduction
= - ( - 0.13 V ) = + 0.13 V
reduction
Br2(l) + 2 e- → 2 Br-(aq) ε⁰ reduction = +1.07 V
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction = + 0.13 V + 1.07 V = 1.20 V
Answer:
0.111 mol
Explanation:
We can solve this problem by using<em> Avogadro's law</em>, which states that at constant pressure and temperature:
Where in this case:
We <u>input the data given by the problem</u>:
- 150.0 mL * n₂ = 50.0 mL * 0.332 mol
And <u>solve for n₂</u>:
I'm 99.99% sure it's false... Hope this helps
