<span>Forward & falling. Hope this helps!</span>
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g
% LI = 13.88/73.89*100 = 18.78% perfectly correct.
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
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Helium only possesses two valence electrons, while the other noble gasses posses eight
