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3 years ago

Why are fluorine (F), chlorine (Cl), and iodine (I) in the same group of the periodic table?

Chemistry

2 answers:

JulsSmile [24]3 years ago

5 0

Answer:

Explanation:

Verdich [7]3 years ago

3 0

Fluorine (F), Chlorine (Cl) and Iodine (I) are all found in the same group on the Periodic Table because they have similar physical properties. Since they are all Halogens, they have 7 valence electrons in their outer shell. In order to get a total of 8, they naturally combine with elements of the same isotope (itself), so D comes close to being correct, but it's not the best answer choice.

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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

How many grams are in 4.07x10^15 molecules of calcium hydroxide

maksim [4K]

There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults

5 0

1 year ago

When a baseball bat hits a baseball, what happens to the energy?<br> completo ancur

professor190 [17]