Soft light (answer) is the filament between (2700k-3000k).
The higher the kelvin number the whiter the light.
3500k-4100k is bright white/cool white
5000k-6500k is daylight
\and those are the three primary colors of color temperature
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Vol.250 before its to much pressure
Answer:
Hi, even though this is really not a question or an answer to a question.
Explanation:
Do you know if any of this stuff involves 6th grade?
Answer:
the ion present in the original solution is Ca2+
Explanation:
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate.
<u>Step1</u> : If we add Nacl to the solution, there is no precipitate formed
⇒The only possible ion that can form a precipate with Cl- is Ag+; since there is no precipitate formed, Ag+ is not present
<u>Step2</u> : If we add Na2SO4 to the solution, a white precipitate is formed
The possible ions to bind at SO42- are Ca2+ and Fe2+
But the white precipitate formed, points in the direction of Ca2+
⇒This means calcium is present
<u>Step3</u> : If we add Na2CO3 to the filtered solution, there is a precipate formed
Ca2+ will bind also with CO32- and form a precipitate
So the ion present in the original solution is Ca2+
