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xeze [42]
3 years ago
13

Identify the Lewis acid in this balanced equation:

Chemistry
2 answers:
VikaD [51]3 years ago
7 0

Answer:

SnCl4

Explanation:

mestny [16]3 years ago
6 0

Answer:

Sncl4

Explanation:

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How many moles of oxygen are in 5.6 moles of al(oh)3?
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5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H

16.8 mols of oxegyn in 5.6 mols of Al(OH)3
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3 years ago
PLEASE HELP ME DUDE MY CLASS ENDS IN LIKE HALF A DAY
mestny [16]

Answer:

c.all atoms have the same arrangement for electrons

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The N H 2 group makes the molecule an amine. Like alcohols, amines can participate in hydrogen bonding. If the drawn molecule en
lesantik [10]

Answer:

a hydrogen bond between the hydrogen of water and the nitrogen of the amine

a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group

Explanation:

A hydrogen bond is formed between molecules in which hydrogen is bonded to a highly electronegative element.

In amines, hydrogen is bonded to nitrogen while in water, hydrogen is bonded to oxygen. Both are highly electronegative elements hence hydrogen bonding is possible between amines and water.

This hydrogen bond may involve;

The hydrogen of water and the nitrogen of the amine

Or

The oxygen of water and a hydrogen from the -NH2 group

7 0
3 years ago
Discuss sucrose transport in vascular plants
Xelga [282]

Answer:Sugars are actively transported from source cells into the sieve-tube companion cells, which are associated with the sieve-tube elements in the vascular bundles. ... The co transport of a proton with sucrose allows movement of sucrose against its concentration gradient into the companion cells.

Explanation:

4 0
3 years ago
G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains \frac{0.6485}{45}×1000 = 14.411 gm-mole. Thus the solubility product K_{sp}of AgBr = [Ag^{+}]×Br^{-}.

Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

Thus S = (5.00×10^{-13})^{1/2} = 7.071×10^{-7} g/mL.

Thus the concentration of Br^{-1} or HBr is 7.071×10^{-7} g/mL.

4 0
4 years ago
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