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Bad White [126]
3 years ago
14

A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

t determine the concentration of the HCl to be if the end point was reached after 50.0 mL of 0.1 M NaOH was added.
Chemistry
1 answer:
irakobra [83]3 years ago
8 0

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

<em>0.00500 moles HCl</em>

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>

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What is the coefficient of silver in the final, balanced equation for this reaction?
vekshin1

This is an incomplete question, the complete question is attached below.

Answer : The coefficient of silver in the final, balanced equation for this reaction is, 3

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given redox reaction is,

Ag^+(aq)+Al(s)\rightarrow Ag(s)+Al^{3+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Al\rightarrow Al^{3+}+3e^-

Reduction : Ag^{+}+1e^-\rightarrow Ag

In order to balance the electrons, we multiply the reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

The balanced redox reaction will be,

3Ag^+(aq)+Al(s)\rightarrow 3Ag(s)+Al^{3+}(aq)

From the balanced redox reaction we conclude that, the coefficient of silver in the final balanced equation for this reaction is 3.

Hence, the correct option is 3.

8 0
4 years ago
Balance this eqaution:<br>​
klasskru [66]

Answer:

H₂ + CuO → Cu + H₂O

Explanation:

We want to balance the given equation;

H₂ + CuO → Cu(OH)₂ + H₂O

To balance this we have to make sure the number of each element on the left is equal to the number of each element on the right.

Thus, the correct balanced equation is;

H₂ + CuO → Cu + H₂O

5 0
2 years ago
A 1.00 L buffer solution is 0.112 M in acetic acid and 0.112 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH
topjm [15]

Answer:

ΔpH = 1.25

Explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.112] / [0.112]

<em>pH = 4.74</em>

The reaction of sodium acetate (CH₃COONa) with HCl is:

CH₃COONa + HCl → CH₃COOH + NaCl

<em>Producing acetic acid, </em>CH₃COOH.

If 0.1mol of HCl reacts the final moles of CH₃COONa are:

0.112mol - 0,1 mol = 0.012mol

Moles of acetic acid are:

0.112mol + 0,1 mol = 0.212mol

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.012] / [0.212]

<em>pH = 3.49</em>

<em></em>

Change in pH, ΔpH = 4.74 - 3.49 =<em> 1.25</em>

I hope it helps!

6 0
3 years ago
Air is made up of about 21% oxygen and 78% nitrogen.
gladu [14]
I believe its A "Air is made up of about 21% oxygen and 78% nitrogen. In this solution, oxygen is the solute and nitrogen is the solvent.
6 0
4 years ago
For a particular isomer of C 8 H 18 , the combustion reaction produces 5099.5 kJ of heat per mole of C 8 H 18 ( g ) consumed, un
xz_007 [3.2K]

Answer:

the standard enthalpy of formation of this isomer of C₈H₁₈ (g) =  -375 kj/mol

Explanation:

The given combustion reaction

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O   ............................(1)

Heat of reaction or enthalpy of combustion = -5099.5 kj/mol

from equation (1)

     ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation                    of reactants)

Or,     - 5099.5   = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)].................................(2)

Given ΔH⁰f(CO₂) = - 393.5 kj/mol  & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0

Using equation (2)

ΔH⁰f(C₈H₁₈) = -621 kj/mol

3 0
3 years ago
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