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3 years ago

Ways to manage non-biodegradable waste

Chemistry

1 answer:

Lisa [10]3 years ago

7 0

Answer:

The following are some ways of safely disposing of non-biodegradable waste.

1. Take them to a local recycling facility. Separate metal, plastic, glass, and other non-biodegradable waste and store them ready for recycling.

2. Combustion and energy recovery.

3. Disposal at sanitary facilities.

4. Reduce.

5. Landfills.

Explanation: Hope it helps! ^_^"

(G00gle)

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3 years ago

Which type of chemical reaction occurs in 2KCIO3 → AKCI + 302?

Oksanka [162]

Answer:

Decomposition reaction

Explanation:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

The given reaction is decomposition reaction.

Decomposition reaction:

It is the reaction in which one reactant is break down into two or more product.

AB → A + B

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

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8 0

3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

Distinguish a substance from a mixture. give two examples of each

amm1812

substance is matter of particular or definite chemical constitution ex: iron, methane
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4 0

3 years ago

Which of the following is not an example of a pure element?

motikmotik

Copper wire is not an example of a pure element because although it's made by pure elements, it's not one itself. It's made by factories.

Have a nice day! :)

8 0

3 years ago

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Ways to manage non-biodegradable waste​

Ways to manage non-biodegradable waste