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Afina-wow [57]
3 years ago
12

Classify the following substances as a Bronsted-Lowry acid, Bronsted-Lowry base, Lewis acid, and/or Lewis base.A. HCl, BF_3,B. C

Cl_3, -HC. CH_2O, CH_3ClD. -OCH_3 , NH_3
Chemistry
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

A. HCl -> Bronsted-Lowry acid, BF3 -> Lewis acid

B. CCl3 -> Lewis acid, -HC -> Lewis acid or Bronsted-Lowry acid

C. H2O -> Bronsted-Lowry acid or Bronsted-Lowry base, CH3Cl -> Lewis base

D.-OCH3 -> Lewis base, NH3 -> Bronsted-Lowry acid or Lewis base.

Explanation:

For the Bronsted-Lowry theory, acids are the substances that can donate a proton H+ and bases are the substances that can receive a pronto H+.

For the Lewis theory, acid is the substance that can gain a pair of electrons, and the base is the substances that can donate the pair of electrons.

A. HCl -> The substance has a proton (H+) and it can donate it, so it's a Bronsted-Lowry acid. BF3 -> The boron (B), still has space in its shells to receive a pair of electrons, so it's a Lewis acid.

B. CCl3 -> The carbon didn't make all the bonds it can do (4), so it still can receive electrons, thus it's a Lewis acid. -HC-> It can lose the proton (H+) as a Bronsted-Lowry acid, os gains a pair of electrons at the carbon, as a Lewis acid.

C. H2O -> It can gain a proton and forms the ion H3O+, or it can lose a proton and form the ion OH-, so it can be a Brosted-Lowy acid or a Bronsted-Lowry base. CH3Cl -> It can donate pair of electrons (the hydrogen is to attached to the carbon, so it will not be lost), so it works as a Lewis base.

D. -OCH3 -> The oxygen still has pairs of electrons, which it can donate, so it's a Lewis base. NH3 -> It can gain a proton and forms the ion NH4+, so it'll be a Bronsted-Lowry acid, but the nitrogen still has electrons, which it can donate, working as a Lewis base.

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3 years ago
You need to prepare 2.00 L of 0.100 M Na2CO3 solution. The best procedure is to weigh out: ___________
jeka94

Answer:

The answer to your question is letter C

Explanation:

Data

Volume = 2 L

Molarity = 0.100 M

Molecular weight Na₂CO₃ = (2 x 23) + (1 x 12) + (3 x 16)

                                           = 46 + 12 + 48

                                           = 106 g

Process

1.- Calculate the grams of Na₂CO₃ needed

                         106 g ----------------  1 mol

                           x      ----------------  0.1 moles

                           x = (0.1 x 106) / 1

                           x = 10.6 g

2.- Calculate the grams of Na₂CO₃ needed for 2 liters of solution

                           10.6 g -------------- 1 liter

                            x        --------------  2 liters

                           x = (10.6 x 2) / 1

                           x = 21.2 grams of Na₂CO₃                        

                     

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