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3 years ago

Why does distillation produce pure water but filtration does not?

1 answer:

7 0

Filtration is used to separate large particles but this process does not provide pure solutions as some impurities still remain in the solution. On the other hand, distillation helps in the formation of pure water as it removes the impurities from the water or solution.

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How many ounces of silver in a silver quarter?

liraira [26]

Each silver quarter has .18 troy ounces of silver. So, my calcs say about 5 1/2 quarters to the troy ounce (silver).

6 0

2 years ago

Some nitrogen for use in synthesizing ammonia is heated slowly, maintaining the external pressure close to the internal pressure

swat32

Answer: The work done for the given process is -2188.7 J

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 50.0 atm

$V_1$ = initial volume = 542 L

$V_2$ = final volume = 974 L

Putting values in above equation, we get:

$W=-50.0atm\times (974-542)L=-21600L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $-21600L.atm=-21600\times 101.33=-2188728J=-2188.7kJ$

The negative sign indicates the system is doing work.

Hence, the work done for the given process is -2188.7 J

7 0

2 years ago

Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allow

aalyn [17]

Answer:

$\Delta G^{\circ}=-15902 J/mol$

Explanation:

In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

$\Delta G^{\circ}=-R*T*ln(K_{eq})$

Being Keq:

$K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}$

Initial conditions:

$[Fructose-1-P]=0.2M$

$[Fructose]=0M$

$[Pi]=0M$

Equilibrium conditions:

$[Fructose-1-P]=6.52*10^{-5}M$

$[Fructose]=0.2M-6.52*10^{-5}M$

$[Pi]=0.2M-6.52*10^{-5}M$

$K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}$

$K_{eq}=613.1$

Free-energy for T=298K (standard):

$\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)$

$\Delta G^{\circ}=-15902 J/mol$

3 0

3 years ago

What did sewage treatment in Ancient Rome and 19th century London have in common?

dolphi86 [110]

Answer:

The history of water supply and sanitation is one of a logistical challenge to provide clean water and sanitation systems since the dawn of civilization. Where water resources, infrastructure or sanitation systems were insufficient, diseases spread and people fell sick or died prematurely.

Explanation:

5 0

3 years ago

The students used 90.0 g/mol as the molar mass of oxalic acid. Use this value to determine how the students calculated the numbe

Svetradugi [14.3K]

Answer:

Hi there, the question asked is not complete but not to worry, I will give an explanation that you will be able to solve similar question or the same question when you get the whole question.

Explanation:

Oxalic acid is a diprotic acid and the acid is used with or react with sodium Hydroxide, NaOH in order to determine the molar mass of unknown diprotic acid. Oxalic Acid reacts with sodium as it is given in the balanced chemical reaction below:

(COOH)₂ + 2NaOH ------------------------------------------------------> Na₂C₂O₄ + 2H₂O.

The amount of the oxalic needed is known, say x gram and the molar mass is known. Thus, the number of moles of the oxalic acid= mass/molar mass = x gram/ 90.0 g/mol.

So we say that (COOH)₂ is a primary standard acid which is been used to standardize the base that is Sodium hydroxide.

Then, if we have an unknown diprotic acid, say H₂A, we can react it with NaOH to get the molar mass of the unknown.

H₂A + 2NaOH ---------------------------------------------------------------> Na₂A + 2H₂O.

Therefore, the number of moles of the acid = [concentration of NaOH × volume of NaOH] × [ 1 mole of the acid/ 2 mole of NaOH].

5 0

3 years ago