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3 years ago

As a student , what can you do to help in preventing pollution in your community ?

Chemistry

1 answer:

aliya0001 [1]3 years ago

3 0

Answer: as a student , i will do to help in preventing pollution in my community :

Reduce the number of trips you take in your car.

Reduce or eliminate fireplace and wood stove use.

Avoid burning leaves, trash, and other materials.

Avoid using gas-powered lawn and garden equipment.

Using non-toxic or less toxic chemicals as cleaners, degreasers and other maintenance chemicals. Implementing water and energy conservation practices. Reusing materials such as drums and pallets rather than disposing of them as waste.

Store your household chemicals (including cleaning products, paints, fuel canisters, etc.) as far away as possible from the living space and, if possible, in metal cabinets. Do not store chemicals in areas where you spend most of the time. Also, do not store chemicals in open cabinets or shelves.

Use Public Transportation. Use your vehicle a lot less often. .

Drive Smart. .

Do Regular Car Check-Up. .

Keep Car Tires Properly Inflated. .

Buy Energy-Efficient Vehicles. .

Consider “going green” .

Plant a Garden. .

Use Low-VOC or Water-based Paints.

Avoid smoking indoors (but quitting smoking is the best answer for overall health).

Use craft supplies in well-ventilated areas.

Make sure your gas stove is well-ventilated.

Minimize clutter.

Remove carpeting if possible.

Use a dehumidifier and/or air conditioner to reduce moisture.

How can students help reduce pollution?

Help Recycling

The most obvious things to reuse are plastic, glass, metal, and paper or fabric (paper and organic fabric are made of the same component, cellulose, and are recycled in a similar way). The color-coded trash bins show where each of these components can be put to be recycled in a proper way.

REDUSE ,REUSE , RECYCLE

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The fluid-filled area between the pleural layers is the.

Snowcat [4.5K]

The fluid filled area between the pleural layers is called the pleural cavity.

8 0

2 years ago

Can anyone explain it to me, please?

Anuta_ua [19.1K]

Answer:

2×10² cg

Explanation:

We'll begin by converting each of the above to the same unit of measurement.

In this case, we shall convert each of the above to kg. This can be obtained as follow:

Conversion of cg to kg

1 cg = 1×10¯⁵ Kg

Therefore,

2×10² cg = 2×10² × 1×10¯⁵

2×10² cg = 0.002 kg

Conversion of dg to Kg

1 dg = 1×10¯⁴ kg

Therefore,

10 dg = 10 × 1×10¯⁴

10 dg = 0.001 kg

Conversion of mg to kg.

1 mg = 1×10¯⁶ Kg

Therefore,

2×10² mg = 2×10² × 1×10¯⁶

2×10² mg = 0.0002 kg

Conversion of ng to kg

1 ng = 1×10¯¹² kg

Therefore,

1×10⁵ ng = 1×10⁵ × 1×10¯¹²

1×10⁵ ng = 0.0000001 Kg

Summary

1. 2×10² cg = 0.002 kg

2. 10 dg = 0.001 kg

3. 2×10² mg = 0.0002 kg

4. 0.001 kg

5. 1×10⁵ ng = 0.0000001 Kg

From the above calculation, 2×10² cg is the highest mass.

5 0

3 years ago

How many moles of C are in a sample of C5H12 that also contains 22.5g of H?

pishuonlain [190]

The number of moles of C present in C₅H₁₂ that contains 22.5 g of H is 9.375 moles

<h3>How to determine the mass of C₅H₁₂ that contains 22.5 g of H</h3>

1 mole of C₅H₁₂ = (12×5) + (1×12) = 72 g

Mass of H in 1 mole of C₅H₁₂ = 12 × 1 = 12 g

Thus,

12 g of H is present in 72 g of C₅H₁₂

Therefore,

22.5 g of H will be present in = (22.5 × 72) / 12 = 135 g of C₅H₁₂

<h3>How to determine the mole of C present in 135 g of C₅H₁₂</h3>

72 g of C₅H₁₂ contains 5 moles of C

Therefore,

135 g of C₅H₁₂ will contain = (135 × 5) / 72 = 9.375 moles of C

Thus, 9.375 moles of C is present in C₅H₁₂ that contains 22.5 g of H

Learn more about mole:

brainly.com/question/13314627

#SPJ1

8 0

2 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

$moles=\frac{1000 grams*1 mole}{78 grams}$

moles=12.82

<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>

5 0

3 years ago

Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard thermodynamic data at 298K, calculate the entropy change f

Ulleksa [173]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4HCl(g)+O_2(g)\rightarrow 2H_2O(g)+2Cl_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cl_2(g))})+(2\times \Delta S^o_{(H_2O(g))})]-[(4\times \Delta S^o_{(HCl(g))})+(1\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(O_2(g))}=205.14J/K.mol\\\Delta So_{(HCl(g))}=186.91J/K.mol\\\Delta S^o_{(Cl_2(g))}=223.07J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (223.07))+(2\times (188.82))]-[(4\times (186.91))+(1\times (205.14))]\\\\\Delta S^o_{rxn}=-129J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-129) J/K = 129 J/K

We are given:

Moles of HCl gas reacted = 2.27 moles

By Stoichiometry of the reaction:

When 4 mole of HCl gas is reacted, the entropy change of the surrounding will be 129 J/K

So, when 2.27 moles of HCl gas is reacted, the entropy change of the surrounding will be = $\frac{129}{4}\times 2.27=73.21J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

3 0

3 years ago

As a student , what can you do to help in preventing pollution in your community ?​

As a student , what can you do to help in preventing pollution in your community ?