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3 years ago

Sulfur does not dissolve in water but it does dissolve in carbon disulfide

Chemistry

2 answers:

Aleks [24]3 years ago

8 0

Answer:

True.

Explanation:

Hello,

Solubility is strongly dependent on both solute's and solvent's polar nature, in such a way, since water is highly polar and sulfur is non-polar, a non-polar solvent must be used to dissolve it. In this case carbon disulfide is proposed, so the way to substantiate it is non-polar is via the difference in the electronegativity of the carbon and sulfur as shown below:

$\Delta E=E_S-E_C\\\Delta E=2.58-2.55\\\Delta E=0.03$

Since such difference is very close to zero making it a highly non-polar substance, consequently, sulfur will dissolve into it.

Best regards.

Lynna [10]3 years ago

3 0

It is really difficult to dissolve the sulfur substance because not only is it polar, but it is composed of long S-chains and not only atoms. So, water cannot dissolve the sulfur because nonpolar compounds do not dissolve in polar solvents. Sulfur doesn't always dissolve with nonpolar solvents, as well. However, since carbon disulfide also contains S-chains, it is the best solvent that would dissolve sulfur.

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3 years ago

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lidiya [134]

Data:
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H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
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Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
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Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

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