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:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
You can calculate potential energy by:
U = m.g.h
Where, U = potential energy
m = mass
g = acceleration due to gravity
h = height
Hope this helps!
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.
Answer:
That right there is what caused the world to end my friend
Explanation:
