3NaOH + FeCl3 → 3NaCl + Fe(OH)3
Charge will decreases.
A parallel plate capacitor when it is fully charged to voltage V is given as:
C = Q/V
The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is
C = ε₀ A /d
since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.
So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.
Thus, Charge will decrease.
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It's mostly used in CHEMICAL PROCESSES.
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
the effect of pressure on surface tension can be attributed in part to absorption of gas at the surface of the liquid and in part to an intrinsic decrease in density of the liquid in the neighborhood of the surface.
In the case of liquids , Owing to contact forces between the edge of the surface and the vessel, the surface acquires a curvature, and if the liquid rises up at the edges where it meets the vessel, the pressure will be less in the liquid than in the air, for points just below and just above the surface. The vessel exerts an upward force on the liquid. This is simply a matter of looking at the directions of forces acting, knowing that the surface is under tension.