Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
Combustion reaction
Key: O2
O2 is normally in a chemical formula when you are used to burn anything, so basically, anything with O2 involves burning.
The reaction equation is:
<span>2CuO(s) + C(s) </span>→ <span>2Cu(s) + CO</span>₂<span>(g)
First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10</span>⁵ g
We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles
Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required
The mass of carbon is then:
5,717 x 12 = 68,604 grams
The mass of coke is:
68,604 / 0.95 = 72,214 g
The mass of coke required is 7.22 x 10⁴ grams
Answer:
concave lens
Explanation:
it's concave lens because it diverges the ray/beam of light.
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles = 
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles = 
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require 
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
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