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Alecsey [184]
3 years ago
12

I need help. Will mark brainliest

Chemistry
2 answers:
Ugo [173]3 years ago
8 0

Answer:

17. HC₂H₃O₂ + H₂O  ⇄  C₂H₃O₂⁻  +  H₃O⁺       Ka

          A             B             CB             CA              

18. H₂SO₄ + 2H₂O →  2H₃O⁺  +  SO₄⁻²

         A           B            CA           CB          

19. CO₃⁻² + H₂O ⇄  HCO₃⁻  +  OH⁻      Kb

      B             A          AC          BC  

20.  HCO₃⁻  + NH₃  ⇄  CO₃⁻²  +  NH₄⁺       Ka

           A           B           BC           AC

21.   NH₄⁺ + OH⁻   ⇄   NH₃  +  H₂O     Ka

         A         B            BC        AC  

Explanation:

The acetic acid release a proton to water, to produce acetate and hydronium.

This is a weak acid.

The sulfuric acid release 2 protons to water to produce sulfate and hydronium.

This acid is considered as strong, but it is only strong in the first dissociation. The second dissociation is weak, with a Ka.

Carbonate takes a proton from water, to become bicarbonate (In this case, the carbonate behaves as a base).

Bicarbonate release a proton to ammonia, to make ammonium and carbonate anion.

Amonium cation release a proton to the hydroxide to make ammonia and water.

exis [7]3 years ago
5 0

Answer:

See explanation

Explanation:

Step 1: Data given

Bronsted-Lowry says the following:

Acid dissociation: HX + H2O → H3O+ + X-

Base dissociation: B + H2O → OH- + HB+

HC2H3O2 + H2O →

HC2HO2 is acetic acid, it's a weak acid. (CH3COOH)

This will be following the reaction HX + H2O → H3O+ + X-

CH3COOH + H2O → CH3COO- + H3O+

H2SO4 + H2O →

H2SO4 is a strong acid

This will be following the reaction HX + H2O → H3O+ + X-

H2SO4 + H2O → HSO4- + H3O+

CO3^2- + H2O →

CO3^2_ is a conjugate base

This will be following the reaction B + H2O → OH_ + HB+

CO3^2- + H2O → HCO3- + OH-

HCO3- + NH3 →

HCO3- (known as bicarbonate) is the conjugate base of H2CO3, a weak acid, and the conjugate acid of the carbonate ion.

NH3 is a weak base

HCO3- + NH3 → CO3^2- + NH4+

NH4+ + OH- →

The hydrogen on the ammonium ion (NH4+) can go back to the hydroxide ion (OH-) to form NH3 and H2O (ammonia and water) again. In this case, because the ammonium ion is donating a proton, it is called a conjugate acid.

NH4+ + OH- → NH3 + H2O

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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
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7 0
3 years ago
A sample has a pH of -0.248 (yes, it really is possible to have a negative pH). Find
dalvyx [7]

Answer:

pOH= 14.248

[H+]=1.77 M

[OH-]=5.65 x10^-15M

Explanation:

pH+pOH= 14

pOH= 14-pH

pOH=14-(-0.248)

pOH= 14.248

[H+]=10^-pH= 10^-(-0.248)=1.77 M

[OH-]=10^-pOH= 10^-14.248=5.65 x10^-15M

5 0
3 years ago
If 5.12 liters of a 2.75 M phosphoric acid is neutralized by magnesium hydroxide solution of 4.00 M.
Tresset [83]

Answer:

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

Molarity of base (Mb) = 4M

Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 3

Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

Vb = 5.28 L

Therefore, the volume of the base is 5.28 L

8 0
4 years ago
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