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Lapatulllka [165]
3 years ago
11

(3) if an aqueous solution of hno3 is electrolyzed for 34.00 min at a steady current of 1.83 a, what volume of h2 (g) at 25.0°c

and 1.05 atm will be collected at the cathode?
Chemistry
1 answer:
tankabanditka [31]3 years ago
3 0
When 2H+  + 2e-  →  H2(g)

when the total charge transfer = steady current * time
                                                    =1.83 * 34 min * 60s = 3733.2 C
and when each electron has 1.6022x10^-19 C
So the amount of charge = 3733.2 / 1.6022x10^-19
                                           = 2.33x10^22 electrons 
To get the moles of the electrons we divided by Avogadro's number
= 2.33x10^22 / 6.022x10^23 = 0.03869 moles
when 2mol e → 1 mol H2
So      0.03869→ ???
moles of H2 = 0.03869 /2 = 0.01935 moles
know we have n = 0.01935 & p = 1.05 atm & R= 0.0821 & T = 25+273=298 K

So by substitution in the ideal gas formula, we can get V:

PV = nRT
1.05 * V = 0.01935 * 0.0821 * 298
∴ V  = 0.45 L



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The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

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3 years ago
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4 years ago
Muscle physiologists study the accumulation of lactic acid [ch3ch(oh)cooh] during exercise. food chemists study its occurrence i
Drupady [299]
The provided information are:
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
 The equation for dissociation of lactic acid is:
CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log \frac{[conjugated base]}{[Acid]}
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86 
The number of moles of lactic acid and lactate are as follows:
n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log \frac{0.45 M}{0.29 M} = 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05
7 0
3 years ago
A 28.5 gram piece of iron is added to a graduated cylinder containing 45.5 mL of water. The water in the cylinder rises to the 4
yan [13]

7.91 g/ml is the density of the iron piece of 28.5 gms.

Explanation:

The density of a substance is defined as the volume it occupies.  It tells the matter present in a substance.

The density is mass per unit volume and is denoted by p.

The formula for density is given by:

density (p) = \frac{mass}{volume}

Data given is :

mass= 28.5 grams

V1 = 45.5 ml

V2= 49.1 ml

The initial volume of water was 45.5 ml, when iron piece of 28.5 grams was added the final volume was 49.1 ml.

Putting the values in the equation of density

p = \frac{28.5}{49.1-45.5}

p = 7.91 g/ml

Since iron is a dense material it will occupy less volume

4 0
3 years ago
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