The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
read more: brainly.com/question/23982245?referrer=searchResults
Answer:
16.5 dm³
Explanation:
Data Given:
no. moles of O₂ = 0.735 moles
volume of O₂ = ?
Solution:
Now
we have to find volume of O₂ gas
Formula used for this purpose
No. of moles = Volume / molar volume
where
molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol
No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)
Put values in equation 1
0.735 = Volume of O₂ / 22.4 dm³/ mol
rearrange above equation
Volume of O₂ = 0.735 x 22.4 dm³/ mol
Volume of O₂ = 16.5 dm³
So,
the volume of O₂ at STP is 16.5 dm³
Answer:
no tree shrubs rainfall is low contains grasses and woody plants correct me if I'm wrong
Answer:
68g/mol
Explanation:
The formula of ammonium sulfide is:
Ammonium sulfide = (NH₄)₂S
The molar mass of a compound is the mass in gram of one mole of the substance. In a compound, it is expressed gram formula mass or gram-molecular weight.
It is determined by the addition of the component atomic masses and then expressed in grams;
Atomic mass of N = 14, H = 1 and S = 32
(NH₄)₂S = 2[14 + 4(1)] + 32 = 36 + 32 = 68g/mol
