2Fe(s) + O2 -> 2FeO(s)
<span>2 'Fe' atoms on both sides </span>
<span>2 'O' atoms on both sides</span>
Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N
Answer: 2.83 J/mol
Explanation:
Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution
Given
No of moles of CaCl = 7.5 mol
Total energy used = 21.2 J
Heat of solution = q/n where
q = total energy
n = number of moles
Heat of solution = 21.2 / 7.5
Heat of solution = 2.83 J/mol
Answer:
As the velocity of light is constant so the acceleration of the light is equal to zero.
a= dv/dt
Explanation:
