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3 years ago

Ability to conduct electricity

Chemistry

1 answer:

Alex17521 [72]3 years ago

8 0

Metals have the ability to conduct electricity.

Metals are lustrous, malleable, ductile and are good conductor of heat and electricity.

They conduct electricity because the electrons in metal are delocalized electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.

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Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

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The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

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3 years ago

(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n

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Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

$5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O$

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene = $\frac{0.03 g}{78 g/mol}=0.0003846 mol$

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

$\frac{15}{5}\times 0.0003846 mol=0.0011538 mol$ of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

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So the unknown solution may have lesser concentration than the isotonic solution so that molecules of that solution move from less concentrated side to the more concentrated side, so its level drops.

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