2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Answer:
A group of cells ready to form roots a stem and the first leaves
Answer:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
Answer:
boron (B), germanium (Ge), and tellurium (Te)
Answer
Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror.
Explanation:
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