No you cannot apply, as the circuit is open, there is no flow of charge or current, consequently the potential drop across the resistances would be zero
You can apply KCL (Kirchhoff's Current Law) and KVL (Kirchhoff's Voltage Law) in both AC and DC analysis. It just gets complicated in AC, because now you have to consider capacitive and inductive reactance, phase angle, power factor, etc. Even in a purely resistive circuit, one without capacitors or inductors, you need to consider AC analysis techniques if the frequency is sufficiently high, because of parasitic reactance that is always present. Kirchhoff's laws are the laws of nature for electrical and electronic circuits.
KCL is Potassium Chloride.
0.012mol KCl x (1L/0.25mol KCl) x (1000ml/1L) = 48 mL KCl
I'm guessing you meant KCl or potassium chloride.
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
KCL is Common to both AC and DC. Only the waveform or AC and DC will differ
Yes. Kirchhoff's laws (current and voltage) are the laws of nature for electrical and electronic circuits.
You can apply KCL (Kirchhoff's Current Law) and KVL (Kirchhoff's Voltage Law) in both AC and DC analysis. It just gets complicated in AC, because now you have to consider capacitive and inductive reactance, phase angle, power factor, etc. Even in a purely resistive circuit, one without capacitors or inductors, you need to consider AC analysis techniques if the frequency is sufficiently high, because of parasitic reactance that is always present. Kirchhoff's laws are the laws of nature for electrical and electronic circuits.
KCl
KCL is Potassium Chloride.
moles KCl = ( M solution ) ( V solution in L )moles KCl = ( 2.2 mol KCl / L solution ) ( 0.635 L of solution )moles KCl = 1.397 moles KCl
0.012mol KCl x (1L/0.25mol KCl) x (1000ml/1L) = 48 mL KCl
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
MW KCl = 74.6 g/mol2.39 gKCl * (1 mol KCL/74.6 g KCl)*(1 L solution/0.06 mol KCL) = 0.534 L
I'm guessing you meant KCl or potassium chloride.
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
KCl is soluble in DMF