The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.
An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Explanation:
The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.
- The nuclear pull is the effect of the nucleus pulling and attracting the electrons in orbitals.
- This pull is stronger for inner orbitals and weak on the outer ones.
- The outer orbitals are said to be well shielded from the pull of the nuclear charge.
- Also, based on the quantum theory, electrons in the outer orbitals have higher energies because they occupy orbitals at having higher energy value.
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Answer: 6.75 moles
Explanation:
This is a simple stoichiometry proboe. that I would set up like this:
(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)
That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.
Hope this helps :)
Answer:
CH₄
Explanation:
To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.
Step 1: Determine the mass of the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g
Step 2: Calculate the percent by mass of each element
%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%
%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%
Step 3: Divide each percentage by the atomic mass of the element
C: 74.8/12.01 = 6.23
H: 25.2/1.01 = 24.95
Step 4: Divide both numbers by the smallest one, i.e. 6.23
C: 6.23/6.23 = 1
H: 24.95/6.23 ≈ 4
The empirical formula of the hydrocarbon is CH₄.
