Help!! 10 points!! Please don’t answer if u don’t know!!
2 answers:
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Given
3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,
Required
the new volume
Solution
Combined with Boyle's law and Gay Lussac's law
T₁ = 39 + 273 = 312
T₂ = 16 + 273 = 289
Input the value :
V₂ = (P₁V₁.T₂)/(P₂.T₁)
V₂ = (99 x 3000 x 289)/(45.5 x 312)
or we can write it as:
V₂ = 3000 L x (289/312) x (99/45.5)
Answer:
%age Yield = 85.36 %
Solution:
The Balance Chemical Reaction is as follow,
C₆H₁₂O + Acid Catalyst → C₆H₁₀ + Acid Catalyst + H₂O
According to Equation ,
100 g (1 mole) C₆H₁₂O produces = 82 g (1 moles) of C₆H₁₀
So,
4.0 g of C₆H₁₂O will produce = X g of C₆H₁₀
Solving for X,
X = (4.0 g × 82 g) ÷ 100 g
X = 3.28 g of C₆H₁₀ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (2.8 g ÷ 3.28 g) × 100
%age Yield = 85.36 %
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
