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2 years ago

Your family is driving across the country to the Grand Canyon what type pollution is your car producing

Chemistry

1 answer:

IgorLugansk [536]2 years ago

8 0

Answer:

The correct answer would be Air Pollution, and Mobile Pollution.

Explanation:

Your family is driving across the country to the Grand Canyon, Air Pollution or Mobile Pollution is produced by your car.

Air pollution is the pollution generated by the Gas emission, Chemical emission, Car Emission, Dust, etc. This pollution consists of solid particles and gases. Air Pollution is very dangerous as it directly affects the Ozone layer.

As the car is moving, so the car is generating the Air pollution, which is considered a mobile source, because car is travelling from one place to another and emitting air pollution where ever it is going.

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The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0

vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of $NH_3$ = 17 g/mole

Given,: 28% w/w of $NH_3$ solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml$

Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute $NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles$

$V_s$ = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.65moles}{0.11L}=15mole/L$

Using molarity equation:

$M_1V_1=M_2V_2$

$15\times V_1=0.036\times 1.0\times 10^{3}$

$V_1=2.4ml$

6 0

3 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

2 years ago

Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so

Artemon [7]

2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)

The molar ratio between NaCN : HCN is 2:2 or 1:1

Mass of HCN = 16.7 g

Molar mass of HCN = 1 + 12 + 14 = 27 g/mol

Molar mass of NaCN = 49 g/mol

Therefore, the mass of NaCN is

16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN

Therefore, 30.3 grams of NaCN gives the lethal dose in the room.

5 0

3 years ago

Read 2 more answers

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

3 years ago

Okay help please will give brainlist

pashok25 [27]

Answer:

Explanation:

hope that helped you

3 0

2 years ago