Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O) = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.
The question is incomplete. The complete question is stated below:
Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.
a. Determine the electric potential at corner A.
b. What is the electric potential energy of a +3 µC charge placed at corner A?
Answer / Explanation:
a )V(A) = 1 / 4πe° ( - 5 5x10∧6C / 0.150m + 2x10∧6C / 0.050m )
The answer to the equation above is : = +6.0x10∧4 j/c
b) U(A) = qV(A)= (3.0x10∧6C) (6.0x10∧4 . j/c) =
The answer to the equation above is : =0.18 J
Explanation:
Where V(A) is equivalent to the electric potential
U(A) is equivalent to the electric potential energy
Answer:
1.what I observe.
Explanation:
The dependent variable in an experiment is what is being observed in the experimental procedure.
This variable is the one that is closely tied to the effects originating from changing the independent variables.
- Independent variables are the ones that cause the observation being studied.
- The effects produced and then studied are the dependent variables.
