A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into 
, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as 
* orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>
Answer:
Final [B] = 1.665 M
Explanation:
3A + 4B → C + 2D
Average rection rate = 3[A]/Δt = 4[B]/Δt = [C]/Δt = 2[D]/Δt
0.05600 M/s = 4 [B]/ 2.50 s
[B] = 0.035 M (concentration of B consumed)
Final [B] = initial [B] - consumed [B]
Final [B] = 1.700 M - 0.035 M
Final [B] = 1.665 M
Answer:
Molecular formula: C₂H₄O₂
Empirical formula: CH₂O
Explanation:
40 % C, 6.72 % H and 53.29 % O states the centesimal composition of the compound. These data means that in 100 g of compound we have x grams of a determined element.
We divide the mass by the molar mass of each:
40 g / 12 g/mol = 3.33 moles of C
6.72 g / 1 g/mol = 6.72 moles of H
53.29 g / 16 g/mol = 3.33 moles of O
We can determine rules of three to get, the molecular formula.
In 100 g of compound we have 3.33 moles of C, 6.72 moles of H and 3.33 moles of O; therefore in 60 g (1 mol) we must have
- (60 . 3.33) / 100 = 2 moles of C
- (60 . 6.72) / 100 = 4 moles of H
- (60 . 3.33) / 100 = 2 moles of O
Molecular formula is C₂H₄O₂
Empirical formula has the lowest suscripts; we divide by two, so the empirical formula is CH₂O
Shape and size has no effect on an objects density
