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3 years ago

Is the process by which certain waste materials, such as plastic, glass, aluminum, and paper, are prepared for reuse.

Chemistry

2 answers:

andriy [413]3 years ago

3 0

the process is called recycling.

kolbaska11 [484]3 years ago

3 0

Explanation:

When waste materials are converted into new materials then this process of development of new materials is known as recycling.

This method helps in the prevention of raw materials and it also prevents wastage of many materials.

For example, glass, paper, aluminium etc can be recycled and hence, they can be used again and again.

Therefore, we can conclude that the process by which certain waste materials, such as plastic, glass, aluminum, and paper, are prepared for reuse is known as recycling.

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A sample of gas at 47C and 1.5 pressure occupies a volume of 2.20L. What volume would this gas occupy at 107C and 2.5 pressure?

Brums [2.3K]

Answer:

Explanation:

8 0

2 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

3 years ago

Suppose 0.708g of copper(II) acetate is dissolved in 50.mL of a 46.0mM aqueous solution of sodium chromate.

lisabon 2012 [21]

Answer:

The final molarity of acetate anion in the solution is 0.0046 moles

Explanation:

The balanced equation is

Cu(C₂H₃O₂)₂ + Na₂CrO₄ = CuCrO₄ + 2Na(C₂H₃O₂)

Therefore one mole of Cu(C₂H₃O₂)₂ react with one mole of Na₂CrO₄ to form one mole of CuCrO₄ and two moles of Na(C₂H₃O₂)

Mass of copper (II) acetate present = 0.708 g

Volume of aqueous sodium present = 50 mL

Molarity of sodium chromate = 46.0 mM

Therefore

Number of moles of sodium chromate present = (50 mL/1000)×46/1000 = 0.0023 M

Number of moles of copper (II) acetate present = 181.63 g/mol

number of moles of copper (II) acetate present = (0.708 g/181.63 g/mol) =0.0039 moles

Therefore 0.0039 moles of Cu(C₂H₃O₂)₂ × (2 moles of Na(C₂H₃O₂))/1 Cu(C₂H₃O₂)₂) = 0.00779 moles of Na(C₂H₃O₂)

also 0.0023 moles of Na₂CrO₄ × (2 moles of Na(C₂H₃O₂))/1 Na₂CrO₄) = 0.0046 moles of Na(C₂H₃O₂)

Therefore the Na₂CrO₄ is the limiting reactant and 0.0046 moles of Na(C₂H₃O₂) or acetate anion is formed

7 0

2 years ago

If 18.0 g of CO2 is produced in the reaction of C2H2 with O2 to form CO2 and

Tanya [424]

Answer:

3.68 g

Explanation:

$n_{CO2}$ = 18 ÷ 44 = 0.4

2C2H2 + 5O2 → 4CO2 + 2H2O

2 : 5 : 4 : 2

0.409 (moles)

⇒ $n_{H2O}$ = 0.409 × 2 ÷ 4 = 0.2045 moles

⇒ $m_{H2O}$ = 0.2045 × 18 = 3.68 grams

5 0

2 years ago

A buffer is prepared by adding 20.0 g of sodium acetate to 480 ml of a 0.160 m acetic acid solution. determine the ph of the buf

docker41 [41]

Hello!

The chemical reaction for the dissociation of Acetic Acid is the following:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺

To solve the problem we are going to use the Henderson-Hasselbach Equation, as follows, but first we need to know the pKa:

$pKa=-log(Ka)=-log( 1,8*10^{-5} )=4,74$

$pH=pKa + log( \frac{[CH_3COO^{-} ]}{[CH_3COOH]} )$

$pH=4,74+log( \frac{\frac{20 g CH_3COONa }{0,480 L}* \frac{1 mol CH_3COONa}{82,03 g CH_3COONa} }{0,160 M} )=5,24$

So, the pH of this solution is 5,24

Have a nice day!

8 0

2 years ago