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Delicious77 [7]
3 years ago
14

Which statement is the best description of a chemical bond?

Chemistry
1 answer:
Margarita [4]3 years ago
5 0

The answer Is B;A chemical bond holds atoms together.

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Hi! Can I have someone that is In the lgbtq+ talk to me, I need some advice...
Kisachek [45]

Answer:

what's the matter?  im an ally... and fully support the lgbtq+ so i might have some advice?

Explanation:

4 0
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What soil particle size has the greatest oxygen holding capacity? What soil particle size has the greatest water holding capacit
Vlad [161]
The soil particle size that has the greatest oxygen holding capacity is the clay. Clay<span> is a fine-grained natural rock or soil material that combines one or more </span>clay minerals with traces of metal oxides and organic matter. I hope my answer has come to your help. God bless and have a nice day ahead!
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3 years ago
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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
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL

Putting values in above equation, we get:

0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

Hence, the mass of glucose in final solution is 1.085 grams

4 0
3 years ago
The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)
EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = \frac{20.0 g}{395 g/mol}=0.05063 mol

2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

3 0
3 years ago
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