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lapo4ka [179]
3 years ago
6

Two plastics that are targeted for recycling from household waste are PETE (1) ( plastic soft drink bottles/ peanut butter and s

alad dressing containers) and HDPE (2) ( milk, water, juice containers/ plastic grocery bags). One of the problems of recycling such materials is separating them. Suppose you have been hired to set up a process for separating large quantities (many tons) of waste plastic that is a mixture of PETE and HDPE. Describe how you might perform this separation.
Any help would be much appreciated!
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
7 0
<span>The HDPE recycling items would usually be clear or lighter in color while the PETE ones would be of color so you could separate it by this color patterns.</span>
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A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of
Rainbow [258]

Answer: (a). T = 38.2 °C     (b). V = 1.3392 cm³     (c). ii and iii  

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

      = yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

  = 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

   n  = PV/RT  

   n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor  in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the  volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

4 0
3 years ago
Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.
aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of  Fe2O3 and CO.

Molar mass of  Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

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now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3  = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

mass of CO = 210.42 g

mass in three significant figures = 210. g

5 0
3 years ago
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Explanation:

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