Chemistry, 21.12.2019 00:31 Travon6524

Which one of the following is not a true statement about water purification. group of answer choices
a. water can be desalinated by reverse osmosis
b. water can be desalinated by distillation
c. water can be purified by chlorination
d. water can be purified by nitration
e. water can be purified by adding aluminum sulfate
f. water can be purified by addition calcium hydroxide

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Answer from: cordovatierra16

d) water can be purified by nitration

Explanation:

a) water can be desalinated by reverse osmosis.

Reverse osmosis is an actual industrial process that is used in the removal of salts, mostly NaCl, from sea water.

In reverse osmosis we go against the natural process of osmosis in which a more concentrated solution is equalized which the concentration of a less concentrated one by the movement of solute through a membrane. This is crucial in biological systems.

b) water can be desalinated by distallitation.

This stament is true since when we distill water what we get is pure water as distillate. This process of purification is terribly expensive.

c) water can be purified by chlorination.

Chlorine is added to water to kill germs hence its use for water treament (note its use in swiming pools)

e) water can be purified by adding aluminum sulfate.

In waste water treatment the addition of aluminum sulfate is to coagulate small particles. This process involves the atraction by the aluminum sulfate of small charged particles and contaminants in raw water which settle in the bottom of the liquid liquid.

f) water can be purified by addition of calcium hydroxide.

This treatment is closely related to the flocculation, the difference is that in coagulation the process is based in the attraction towards charged particles whereas flocculation involves binding small particles into bigger ones which in turn attract more particles growing in size.

This leaves option d) as the false one. Nitration actually will make water more acidic and is never used to purify water.

Answer from: Quest

Cr(oh)3 -> cr(3+) + 3oh-

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.