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Nata [24]
3 years ago
8

Titration of 25.0 mL of an HCl solution of unknown concentration requires 14.8 mL of 0.100 M NaOH. What is the molar concentrati

on of the HCl solution
Chemistry
1 answer:
nordsb [41]3 years ago
4 0
M1V1 = M2V2
M1 = 0.100 M NaOH
V1 = 14.8 mL NaOH
M2 = ?
V2 = 25.0 mL HCl

(0.100 M)(14.8 mL) = M2(25.0 mL)

M2 = 0.0592 M HCl
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According to the periodic table, lithium and nitrogen have the same number of
Vaselesa [24]
Lithium and nitrogen have the same electron energy levels, actually 2 levels.
6 0
3 years ago
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Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(b) Hg₂²⁺(aq)
Olegator [25]

The given reaction is not spontaneous.

We must recognize changes in oxidation states that take place across elements in order to balance these equations. To accomplish this, keep in mind following guidelines:

A neutral element on its own has an oxidation number of zero.For a neutral molecule, the total number of oxidations must be zero.The net charge of an ion is equal to the sum of its oxidation numbers.In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.In a compound with no oxygen present the other halogens will also prefer -1.

One of the mercury atoms is oxidized from +1 to +2 in the simple aqueous ion, for a loss of 1 electron.

Oxidation half-reaction:

0.5 Hg^{2+} _{2} (aq) →Hg^{2+} (aq) + 1e^-

E^o _{ox} = - 0.92 V

The other mercury is reduced from +1  to zero in mercury metal, for a gain of 1 electron.

Reduction half-reaction:

0.5 Hg^{2+} _{2} (aq) + 1 e^- →Hg(l)

E^o _{red} = 0.85V

This is a disproportionation redox reaction !

Net reaction:

Hg^{2+} _{2} (aq) →Hg^{2+} (aq) + Hg (l)

E^o _{cell} = 0.85 - 0.92 = -0.07V

The cell potential is negative so this reaction is NOT spontaneous.

To learn more about the non spontaneous reaction please click on the link brainly.com/question/20358734

#SPJ4

5 0
2 years ago
An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI prese
Maru [420]

Answer: The molarity of Na_2S_2O_3 is 0.108 M

Explanation:

KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2

2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}    

\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol

1 mole of KIO_3  produces = 3 moles of   I_2

2.8\times 10^{-4} moles of KIO_3 produces = \frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4} moles of I_2  

Now 1 mole of I_2 uses = 2 moles of Na_2S_2O_3

8.4\times 10^{-4} moles of I_2 uses =  \frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}  moles of Na_2S_2O_3

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M

The molarity of Na_2S_2O_3 is 0.108 M  

6 0
3 years ago
Which reaction creates more massive nuclei?
maxonik [38]

Answer:

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hope this helps

6 0
3 years ago
Which statement is true about elements in a period on the periodic table?
frozen [14]

Answer:

they have differnet properties that repeat across the next period

Explanation:

5 0
3 years ago
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