Answer : The maximum concentration of silver ion is 
Solution : Given,
for AgBr = 
Concentration of NaBr solution = 0.1 m
The equilibrium reaction for NaBr solution is,
The concentration of NaBr solution is 0.1 m that means,
![[Na^+]=[Br^-]=0.1m](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D%5BBr%5E-%5D%3D0.1m)
The equilibrium reaction for AgBr is,
At equilibrium s s
The expression for solubility product constant for AgBr is,
![K_{sp}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
The concentration of 
= s
The concentration of 
= 0.1 + s
Now put all the given values in expression, we get
By rearranging the terms, we get the value of 's'
Therefore, the maximum concentration of silver ion is .
Answer:
605 km
Explanation:
Hello
the same units of measure should be used, then
Step 1
convert 42 m/s ⇒ km/h
1 km =1000 m
1 h = 36000 sec
Step 2
find kilometers traveled after 4 hours
V,velocity
s, distance traveled
t. time
now, isolating s
and replacing
S=604.8 Km
Have a great day
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out.
Cosin (40) is equal to around .766
Adjacent/Hypotenuse
x/120 = cos40
Answer: 91.92533.
If you use 3 significant figures it should be 91.9 meters.
Answer:
I think it's a because it goes thru it and reflects
Answer:
(a) The energy of the photon is 1.632 x 
J.
(b) The wavelength of the photon is 1.2 x 
m.
(c) The frequency of the photon is 2.47 x 
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
- = -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x 
)
- = 1.632 x Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x 
Js), c is the speed of light (3 x 
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x * 3 x )/ λ
λ = 
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x x f
f = 
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.
