Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as

Where
is the quantum number
now
We can look at the given angular momentum in this form as

comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
corrected question:
Determining Density and Using Density to Determine Volume or Mass
(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³
(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.
(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?
(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?
Answer:
density = 
ρ=m/v ,m=ρv, v=m/ρ
(a)m=1*10g , v=7.36cm³
ρ=10/7.36 =1.36g/cm³
(b) m=65g, ρ=0.791 g/mL.
v= 65/0.791 =82.17g/mL
(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³
m=19..32*8=154.56g/cm³
(d) mass of copper=374.5g , v=41.8cm³
ρ=374.5/41.8 =8.96g/cm³
mass of ethanol=15g, density of ethanol=0.789g/mL
v=15/0.789 =19.01mL
volume of mecury=25mL, density of mercury=13.6g/mL
m=25*13.6=340g
Answer
PubChem CID/molecular formula
Explanation:
Cesium bromide
PubChem CID 24592
Molecular Formula CsBr or BrCs
Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...
Molecular Weight 212.81 g/mol
Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)
have a good day /night
may i please have a branllist
It's been a while... but I'm <em>pretty</em> sure it's
H₂O₄
<u>Answer:</u> The time taken by the reaction is 84.5 seconds
<u>Explanation:</u>
The equation used to calculate half life for first order kinetics:

where,
= half-life of the reaction = 9.0 s
k = rate constant = ?
Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
......(1)
where,
k = rate constant = 
t = time taken for decay process = 50.7 sec
= initial amount of the reactant = ?
[A] = amount left after decay process = 0.0741 M
Putting values in equation 1, we get:
![0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}](https://tex.z-dn.net/?f=0.077%3D%5Cfrac%7B2.303%7D%7B50.7%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B0.0741%7D)
![[A_o]=3.67M](https://tex.z-dn.net/?f=%5BA_o%5D%3D3.67M)
Now, calculating the time taken by using equation 1:
![[A]=0.0055M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0055M)

![[A_o]=3.67M](https://tex.z-dn.net/?f=%5BA_o%5D%3D3.67M)
Putting values in equation 1, we get:

Hence, the time taken by the reaction is 84.5 seconds