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3 years ago

10

charcoal removes fine particulates, as well as lead, mercury, copper, and chlorine from water through a separation process calle

d

1 answer:

Pachacha [2.7K]3 years ago

7 0

Absorption distillation floatation filtration

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At what altitude could a climber expect to first see snow?

Andrews [41]

Answer: 600
I hope this helps you

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3 years ago

Which acid could not be prepared by treating a Grignard reagent with CO2?

Anni [7]

Answer:

4-oxopentanoic acid.

Explanation:

In this case, we must remember that the Grignard reaction is a reaction in which <u>carbanions</u> are produced. Carboanions have the ability to react with CO2 to generate a new C-C bond and a carboxylate ion. Finally, the acid medium will protonate the carboxylate to produce the <u>carboxylic acid group. </u>

The molecules that can follow the mechanism described above are the molecules: p-methylbenzoic acid, cyclopentane carboxylic acid and 3-methylbutanoic acid. (See figure 1)

In the case of <u>4-oxopentanoic acid</u>, the possible carbanion <u>will attack the carbonyl group</u> to generate a cyclic structure and an alcohol group (1-methylcyclopropan-1-ol). Therefore, this molecule cannot be produced by this reaction. (See figure 2)

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3 years ago

The density of titanium is 4.51 g/cm^3. what is the volume (in cubic inches) of 3.5 lb of titanium?

Lina20 [59]

<span>(3.5 lb Ti) x (453.592 g Ti / 1 lb Ti) x (1 cm^3 Ti / 4.51 g Ti) x 0.0610237 in^3 / 1 cm^3) = 21 in^3 Ti. Use factor label method to cancel out units and make sure to cancel out the units to the solution, which in this case is volume of titanium in cubic inches. We only keep two digits because the original numbers use have two sig figs as the least amount when doing multiplication or division.</span>

4 0

3 years ago

Read 2 more answers

What is the solubility of argon (in units of grams per liter) in water at 25 °C, when the Ar gas over the solution has a partial

beks73 [17]

Answer:

0.02405 g/L is the solubility of argon in water at 25 °C.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{Ar}=K_H\times p_{gas}$

where,

$K_H$ = Henry's constant = $1.40\times 10^{-3}mol/L.atm$

$p_{Ar}$ = partial pressure of carbonated drink = 0.51atm

Putting values in above equation, we get:

$C_{Ar}=1.40\times 10^{-3}mol/L.atm\times 0.430 atm\\\\C_{Ar}=6.02\times 10^{-4}mol/L$

Molar mass of argon = 39.95 g/mol

Solubility of the argon gas :

$6.02\times 10^{-4}mol/L\times 39.95 g/mol=0.02405 g/L$

0.02405 g/L is the solubility of argon in water at 25 °C.

5 0

4 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

3 years ago