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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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15) The average human is ~60% water, which translates to ~44.3 kg of water. If you eat a Reece's peanut butter cup (105 Calories

EastWind [94]

Answer: D) 2.37°C

Explanation:

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3 0

2 years ago

If the p-side has a higher doping concentration, explain how to keep tuning to the same radio channel?

zzz [600]

Answer:

increase in temperature of the intrinsic semiconductor

Explanation:

If the p-side has a higher doping concentration, it implies that number of holes (positive ion) increased which is greater than number of electron (negative ion) in the n-side
in order to balance the intrinsic concentration, that is to balance the number of holes and electrons which depends on temperature.
an increase in the temperature of the intrinsic semiconductor (p-side), increases the number of electron but number of holes remains constant.

A balance in the intrinsic concentration helps in tuning to the same radio channel.

6 0

4 years ago

Calculate the mass (in grams) of 0.473 mol of titanium

xz_007 [3.2K]

Explanation:

n=given mass ÷molar mass

make given mass become the subject of the formula by

multiplying the molar mass on both sides of the equation.

n=0.473mol

given mass=??

molar mass=48

therefore,given mass=n×molar mass

=0.473×48

=22.704grams

mass in grams is 22.704grams

7 0

3 years ago

What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?

vova2212 [387]

The volume of SO2 produced at 325k is calculated as below

calculate the moles of SO2 produced which is calculated as follows

write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2

find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles

by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2

use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm

V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2

3 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

3 years ago