Answer:
kinetic energy I think so
Since that E°cell =E°cathode - E°anode then
0.62V = 0.34 - E°anode
Therefore E° anode is = -0.28V
We know the volume of one mole of gas at 273 K and 760 Torr is 22.4 L. Using
(PV)/T = constant
We can calculate the volume of the gasses at the given conditions:
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4) / 273 = (288 V₂) / 308.2
V₂ = 66.7 L
Mass of He: 4
Mass of Ne: 20
Fraction of Ne: x
Fraction of He: 1 - x
avg density = (∑(component fraction × component mass))/volume
0.2460 = (20x + 4(1 - x))/ 66.7
x = 0.775
The mass of 
that would be formed will be 18.22 grams
<h3>Stoichiometric calculations</h3>
Let us first look at the balanced equation of the reaction:
The mole ratio of Y to 
is 2:3.
Mole of 10.0 grams of Y = 10/88.9 = 0.11 moles
Mole of 10.0 grams = 10/71 = 0.14 moles
3/2 of 0.11 = 0.165. Thus, is limiting in availability.
Mole ratio of and = 3:2
Equivalent mole of = 2/3 x 0.14 = 0.093 moles.
Mass of 0.093 moles =0.093 x 195.26 = 18.22 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Answer:
It is a base and should turn a paper green
Explanation:
