Answer:
Explanation:
All three lighter boron trihalides, BX3 (X = F, Cl, Br), form stable adducts with common Lewis bases. Their relative Lewis acidities can be evaluated in terms of the relative exothermicities of the adduct-forming reaction. Such measurements have revealed the following sequence for the Lewis acidity: BF3 < BCl3 < BBr3 (in other words, BBr3 is the strongest Lewis acid).
This trend is commonly attributed to the degree of π-bonding in the planar boron trihalide that would be lost upon pyramidalization (the conversion of the trigonal planar geometry to a tetrahedral one) of the BX3 molecule, which follows this trend: BF3 > BCl3 > BBr3 (that is, BBr3 is the most easily pyramidalized). The criteria for evaluating the relative strength of π-bonding are not clear, however. One suggestion is that the F atom is small compared to the larger Cl and Br atoms, and the lone pair electron in the 2pzorbital of F is readily and easily donated, and overlaps with the empty 2pz orbital of boron. As a result, the [latex]\pi[/latex] donation of F is greater than that of Cl or Br. In an alternative explanation, the low Lewis acidity for BF3 is attributed to the relative weakness of the bond in the adducts F3B-L.
Answer:
-3.28 × 10⁴ J
Explanation:
Step 1: Given data
- Pressure exerted (P): 27.0 atm
- Initial volume (Vi): 88.0 L
- Final volume (Vf): 100.0 L
Step 2: Calculate the work (w) done by the gaseous mixture
We will use the following expression.
w = -P × ΔV = -P × (Vf - Vi)
w = -27.0 atm × (100.0 L - 88.0 L)
w = -324 atm.L
Step 3: Convert w to Joule (SI unit)
We will use the conversion factor 1 atm.L = 101.325 J.
-324 atm.L × 101.325 J/1 atm.L = -3.28 × 10⁴ J
Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.
Its b because it explains it better than a waterfall does
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You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =
