C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole
D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles
<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>
1 mole of N₂H₄ contains 4 moles of H
Therefore,
0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H
<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>
1 mole of C₁₀H₂₂ contains 22 moles of H
Therefore,
34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H
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Answer:
Ethanol
Explanation:
2 C atoms and a single OH group.
Formula: C2H5OH
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.
