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Olegator [25]
3 years ago
11

Give the formula of each of the following (don't worry about subscripts or superscripts, example: HCO3- enter as HCO3-): the con

jugate acid of HSO3- H2SO3 the conjugate base of H2O the conjugate acid of NH3 the conjugate base of CH3NH2
Chemistry
2 answers:
azamat3 years ago
8 0

Answer:

See explanation below

Explanation:

In this case, for you to understand, let's do the acid base reaction, so you can see the conjugate base or acid.

For the case of the HSO3-, if you do the reaction:

HSO3- + H2O ------> H2SO3 + OH-

In this case, the negative charge, acts as a base, and took an hydrogen from the water. This forms the OH- and the conjugate acid of the HSO3-. The H2SO3 is called sulfurous acid.

For the case of H2SO3:

H2SO3 + H2O ---------> HSO3- + H3O+

The H2SO3 acts like acid, so the conjugate base would be the HSO3-

For the H2O:

2H2O --------> H3O+ + OH-

The water acts as base and acid at a time. In this case, when it acts like a base, took one hydrogen atom from the other molecule of water, and forms the H3O+. This would be the conjugate acid. The conjugate base would be the OH-

For the NH3:

NH3 + H2O --------> NH4+ + OH-

The atom of nitrogen in the NH3, took an atom of hydrogen to form the NH4+, which would be the conjugate acid.

For the CH3NH2:

CH3NH2 + H2O -------> CH3NH- + H3O+

The water took an atom of hydrogen, so, the water is the base and CH3NH2 is the acid, so the conjugate base is the CH3NH-

Regards

vovangra [49]3 years ago
3 0

Answer:

The relative conjugate acids and bases are listed below:

CH3NH2 → CH3NH3+

H2SO3→ HSO3-

NH3→ NH4+

Explanation:

In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species resulting from a base accepting a proton; likewise, a conjugate base is the species formed after an acid has donated a hydrogen atom (proton).

To this end:

  • HSO3- is the conjugate acid of H2SO3 i.e sulfuric acid has lost a proton (H+)
  • NH4+ is the conjugate acid of NH3 i.e the base ammonia has gained a proton (H+)
  • OH- is the conjugate base of H20
  • CH3NH3+ is the conjugate base of the base CH3NH2 methylamine
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The bromination of acetone is acid-catalyzed.CH3COCH3 + Br2 CH3COCH2Br + H+ + Br -The rate of disappearance of bromine was measu
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Answer:

a) The rate law is:

rate = k[Acetone][Br₂]⁰[H⁺] = k[Acetone][H⁺]

b) The value of k is:

k = 3.86 × 10⁻³ M⁻¹ · s⁻¹

Explanation:

Acetone (M) Br2 (M) H+ (M) Rate (M/s)

0.30                 0.050 0.050 5.7 x 10-5

0.30                   0.10 0.050 5.7 x 10-5

0.30                  0.050    0.10       1.2 x 10-4

0.40              0.050  0.20  3.1 x 10-4

0.40               0.050         0.050 7.6 x 10-5

A generic rate law for this reaction could be written as follows:

rate = k[Acetone]ᵃ[Br₂]ᵇ[H⁺]ⁿ

The rate for the reaction in trial 2 is:

rate 2 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ

For the reaction in trial 1:

rate 1 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

If we divide both expressions, we can obtain "b": rate2 / rate1:

rate2/rate1 = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ / k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

1 = 2ᵇ

b = 0

If we now take the expressions from trial 3 and 1 and divide them, we can obtain "n":

rate 3/rate 1 = k(0.3)ᵃ(0.050)⁰(0.01)ⁿ/ k(0.3)ᵃ(0.050)⁰(0.050)ⁿ

2.1 = 2ⁿ  Applying ln to both side of the equation:

ln 2.1 = n ln2

ln2.1/ln2 = n

1 ≅ n

Taking now the reaction in trial 5 and 1 and dividing them:

rate 5/rate 1 = k(0.4)ᵃ(0.050)⁰(0.050) / k(0.3)ᵃ(0.050)⁰(0.050)

4/3 = 4/3ᵃ  

a = 1

a)Then the rate law can be written as follows:

rate = k[Acetone][Br₂]⁰[H⁺]

It might be suprising that the rate of bromination of acetone does not depend on the concentration of Br₂. However, looking at the reaction mechanism, you can find out why.

b) Now, we can find the constant k for every experiment and calculate its average value:

rate / [Acetone][Br₂]⁰[H⁺]  = k

For reaction 1:

k1 = 5.7 ×10⁻⁵M/s / (0.3 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 2: k2 = 5.7 ×10⁻⁵M/s / (0.30 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 3: k3 = 1.2 ×10⁻⁴M/s / (0.30 M)(0.10 M) = 4.0 ×10⁻³ M⁻¹ · s⁻¹

Reaction 4: k4 = 3.1 ×10⁻⁴M/s / (0.40 M)(0.20 M) = 3.9 ×10⁻³ M⁻¹ · s⁻¹

Reaction 5: k5 = 7.6 ×10⁻⁵M/s / (0.4 M)(0.05 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Averge value of k:

k = (k1 + k2 + k3 + k4 + k5)/5 = 3.86 × 10⁻³ M⁻¹ · s⁻¹

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Answer:

96.09 g/mol

Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

N = 14.01 g/mol

H = 1.008 g/mol

O = 16.00 g/mol

C = 12.01 g/mol

Since your compound is  

(

N

H

4

)

2

C

O

3

, you need to multiply the atomic weights by their subscripts. Therefore,

Step 2

N = 14.01 g/mol × 2 =

28.02 g/mol

H = 1.008 g/mol × (4×2) =

8.064 g/mol

 

O = 16.00 g/mol × 3 =

48.00 g/mol

C = 12.01 g/mol × 1 =

12.00 g/mol

To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

molar mass of

(

NH

4

)

2

CO

3

=

(28.02 + 8.064 + 48.00 + 12.01) g/mol

=

96.09 g/mol

this is a google search and a example i hope is helps to solve

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3 years ago
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