<span>fossil fuels are mainly organic compounds.Combustion of fossil fuels also produces other air pollutants, such as nitrogen oxides, sulfur dioxide, volatile organic compounds and heavy metals.so,substances which contain sulphur cause to form harmful acids like sulphuric and sulphurus acids</span>
<u>Answer:</u> The concentration of the solution is 0.25 M
<u>Explanation:</u>
Let the volume of solution of 2.5 M NaCl be 10 mL
We are given:
Dilution ratio = 1 : 10
So, the solution prepared will have a volume of = 
To calculate the molarity of the diluted solution, we use the equation:
where,
are the molarity and volume of the concentrated NaCl solution
are the molarity and volume of diluted NaCl solution
We are given:
Putting values in above equation, we get:
Hence, the concentration of the solution is 0.25 M
If there is an increase in industrial activity, that means that more heat will be dissipated to the atmosphere in the form of carbon dioxide. Industrialization requires fuel to keep the processes on the go. At the end of the pipeline, the combustion of fuel would result to carbon dioxide released to the atmosphere. That's how it is contributing to the global climate change through the greenhouse effect.
Answer : The correct option is, (C) 2, 4 and 5.
Explanation :
Combustion reaction : It is a type of reaction in which a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.
For example : Methane react with oxygen to give carbon dioxide and water.
In the given list of chemical substances, 
are in oxide form. They can not be both reactant and product of a single combustion reaction.
In the given list, 
is the only hydrocarbon which shows a combustion reaction. That means react with 
to give 
and 
as a product.
The balanced combustion reaction of is,
Therefore, the correct answer is, (C) 2, 4, and 5.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
