Answer:
Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v
Explanation:
(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v
(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v
________________________________________________
Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)
Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)
= 0.34v + 2.37v = 2.72v
Answer:
223.5 g
Explanation:
The formula between the number of moles, mass and Mr can be used to convert moles to grams.
<em>Number of moles = mass ÷ Mr</em>
So, mass = number of moles × Mr
Mr of Li₂O = (6.9 × 2) + 16 = 29.8
∴ Mass = 7.5 × 29.8 = <u>223.5 g</u>
<h3>
Answer:</h3>
43.27 g Mg
<h3>
Explanation:</h3>
The balanced equation for the reaction between magnesium metal and hydrochloric acid is;
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
From the equation;
1 mole of magnesium reacts with 2 moles of HCl
We are given;
3.56 moles of Mg and 3.56 moles of HCl
Using the mole ratio;
3.56 moles of Mg would react with 7.12 moles of HCl, and
3.56 moles of HCl would react with 1.78 moles of Mg
Therefore;
The amount of magnesium was in excess;
Moles of Mg left = 3.56 moles - 1.78 moles
= 1.78 moles
But; 1 mole of Mg = 24.305 g/mol
Therefore;
Mass of magnesium left = 1.78 moles × 24.305 g/mol
= 43.2629 g
= 43.27 g
Thus, the mass of magnesium that remained after the reaction is 43.27 g
Answer:
NaOH + HCl → H2O and NaCl.
Explanation:
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Answer:
C)
Explanation:
It's a covalent bond because O and H are both non metals.
