Answer:
The theoretical yield of the hydrogen sulfide is 68.0 grams.
Explanation:
Moles of aluminum sulfide = 4.0 mol
Moles of water = 4.0 mol
According to reaction, 6 moles of water reacts with 1 mole of aluminum sulfide,then 4 moles of water will react with :
of aluminum sulfide
1.5 moles aluminum sulfide < 4 moles aluminum sulfide
This means that water is present in limiting amount and aluminum sulfide is in excess amount.So, amount of hydrogen sulfide will depend upon moles of water.
According to reaction, 6 moles of water gives with 3 mole of hydrogen sulfide,then 4 moles of water will give :
of hydrogen sulfide
Mass of hydrogen sulfide:
2.0 mol × 34 g/mol = 68.0 g
The theoretical yield of the hydrogen sulfide is 68.0 grams.
Just divide 675 by the atomic number of cesium and you will end up with
5.078798426627975
Answer:
3.25×10²⁴ molecules
Explanation:
From the question given above, the following data were obtained:
Mass of H₂O = 97.2 g
Number of molecule of H₂O =?
From Avogadro's hypothesis, we understood that:
1 mole of H₂O = 6.02×10²³ molecules
Next, we shall determine the mass of 1 mole of H₂O. This can be obtained as follow:
1 mole of H₂O = (2×1) + 16
= 2 + 16
= 18 g
Thus,
18 g of H₂O = 6.02×10²³ molecules
Finally, we shall determine the number of molecules in 97.2 g of H₂O. This can be obtained as follow:
18 g of H₂O = 6.02×10²³ molecules
Therefore,
97.2 g of H₂O = 97.2 × 6.02×10²³ / 18
97.2 g of H₂O = 3.25×10²⁴ molecules
Thus, 97.2 g of H₂O contains 3.25×10²⁴ molecules.
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
