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Vika [28.1K]
2 years ago
6

what additional reaction product would result if naoh was used as a base in this reaction instead of k2co3?

Chemistry
1 answer:
AnnZ [28]2 years ago
7 0

Answer:

What reaction product would result if NaOH was used as a base in this reaction instead of K2CO3? If sodium hydroxide was used as the base of the reaction, then the reaction product formed would be an ethyl alcohol because it would undergo a hydrolysis reaction with addition of iodoethane

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Which of the following are chemicals?
Effectus [21]
I think it is c because food and clothing aren't technically chemicals however they do contain them so d is a possible answer but I think it is c
3 0
3 years ago
The following calculations must be handwritten in your notebook. – Acetic Acid ■ Hydrogen ion concentration ■ Ka ■ % Error – Ace
Nikitich [7]

Answer:

Explanation:

1) Acetic acid

Concentration is given as 0.103 M

The average pH of this solution = 2.96

we know that pH = - log [H+] therefore [H+] = 10-pH

[H+] = 10-2.96

= 1.1 x 10-3 M = 0.0011 M

Consider the equilibrium

CH3COOH ⇄CH3COO- + H+

Initial 0.103 0 0

Change -x +x +x

equlibrium 0.103 -x x x

Ka = x2 / 0.103 - x

Here the initial concentration of CH3COOH = 0.103 M

the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5

2) Acetic acid + NaOH

pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

Volume and conc of acetic acid = 10 mL of 0.103 M

= 10 mL x 0.103 mmol / mL

= 1.03 mmol

Volume and conc of NaOH added = 4 mL of 0.0992 M

= 4 x 0.0992 mmol

= 0.397 mmol

Consider the equation

CH3COOH + NaOH -----------> CH3COONa + H2O

Initial 1.03 0.397 0

Final 0.633 0 0.397

0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

pH = pKa + log 0.627 = pKa - 0.203

or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]

= 4.683

Ka = 10-4.683 = 2.07 x 10-5

3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

x =  1.03 x 10-4

Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99

(ii) Mass of Na2HPO4.7H2O =0.601 g

therefore no of moles = 0.601 / 268.07 = 0.00224 mole

= 2.24 mmol

The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml

= 0.075 M

consider the equilibrium

HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

(iii) Mass of Na3PO4.12H2O taken = 0.208 g

moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

Consider the equilibrium reaction

PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

equilibrium 0.055-x x x

Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]

x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

That is the conce of OH- ion = 0.025M

Therefore pH = 14 - pOH = 14 - 1.6 =12.4

3 0
2 years ago
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of
erik [133]

Answer:

Ba\ percentage\ in\ Mass=4.8\%

Explanation:

From the question we are told that:

Mass of mixture m=3.455g

Mass of Barium m_b=0.2815g

Equation of Reaction is given as

Ba2+ + H2SO4 => BaSO4 + 2 H+

Generally the equation for Moles of Barium  is mathematically given by

Since

 Moles of Ba^{2+} = Moles of BaSO_4

Therefore

 Moles of Ba^{2+}  = \frac{mass}{molar mass of BaSO4}  

 Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol

Generally the equation for Mass of Barium  is mathematically given by

 Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}  

 Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g

Therefore

 Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%    

 Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%

 Ba\ percentage\ in\ Mass=4.8\%

8 0
3 years ago
Please help asap and thank you
natulia [17]

Answer:

Metamorphic rock

Explanation:

metamorphic rocks do not get hot enough to melt they just get denser if they did they would become igneous rocks.

7 0
3 years ago
Which organism is most likely 10 centimeters in size?
djyliett [7]

Answer:

Actually, one of the more interesting organisms at those depths is the Xenophyophore, a creature which, despite being single-celled, can grow to be over 10 centimeters wide. "Scientists say xenophyophores are the largest individual cells in existence.

Explanation:

4 0
2 years ago
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