I'm assuming that by "miles" you mean moles.
If O2 is the excess reactant, that means Fe is the limiting reactant. That means that the amount of product being formed depends on the amount of Fe reactant present. To calculate the moles of Fe2O3 formed, start with the given 6.4 moles of Fe and use the mole to mole ratio given by the reaction as shown below:
6.4 mol Fe x

=
3.2 mol Fe2O3
<u>Answer:</u> The correct answer is Option d.
<u>Explanation:</u>
According to Lewis acid-base concept:
The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.
For the given chemical reaction:

is accepting electron pair and is getting converted to
. Thus, it is considered as Lewis acid.
present in CuO is a Lewis base because it is donating electron pair.
Thus, the correct answer is Option d.
Answer:
4.13×10²⁷ molecules of N₂ are in the room
Explanation:
ideal gases Law → P . V = n . R . T
Pressure . volume = moles . Ideal Gases Constant . T° K
T°K = T°C + 273 → 20°C + 273 = 293K
Let's determine the volume of the room:
18 ft . 18 ft . 18ft = 5832 ft³
We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L
1 atm . 165045.6 L = n . 0.082 L.atm/mol.K . 293K
(1 atm . 165045.6 L) / 0.082 L.atm/mol.K . 293K = n
6869.4 moles of N₂ are in the room
If we want to find out the number of molecules we multiply the moles by NA
6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules
Answer:
because the electron is a negatively charge and have high energy
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.