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2 years ago

11

The reaction described in Part A required 3.40 L of sodium chloride. What is the concentration of this sodium chloride solution

1 answer:

Mumz [18]2 years ago

5 0

Answer:

Part A:

First, convert molarity to moles by multiplying by the volume:

0.293 M AgNO3 = (0.293 moles AgNO3)/1 L x 1.19 L = 0.349 moles AgNO3

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What does standard deviation reveal about data?

NISA [10]

Answer:

A. The average of all the data points

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3 years ago

How many protons does silicon have? <br> a. 2 <br> b. 14 <br> c. 28 <br> d. 28.08

EleoNora [17]

The answers is, B. 14

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2 years ago

What is the number of electrons will chlorine most likely gain/lose

LuckyWell [14K]

Gain 1.

This is because of Chlorine's placement on the periodic table. Chlorine is a Halogen, thus being located in group 17 or 7A. This means it is one of the closest elements to being a Noble Gas or group 18/8A.

Chlorine has 7 valence electrons, and in order for it to become a Noble Gas, it needs 8. Cl- is typically the symbol used to represent this as Chlorine needs to gain an electrons instead of lose; once Chlorine gains the electron, it will be stable.

This also means that Chlorine has a high ionization energy or, in simpler terms, it would be difficult to remove an electron from Chlorine.

Hope this helps!

8 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago

15. You have a solution of .835mol of NaCl in 2.5L of solution, what is the molarity?

Viktor [21]

Answer: 0.334mol/L

Explanation:

$M=\frac{mol}{L}$

$M=\frac{0.835mol}{2.5L} \\M=0.334mol/L$

5 0

2 years ago