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3 years ago

CO is a primary pollutant. a. True b. False?

Chemistry

2 answers:

Y_Kistochka [10]3 years ago

3 0

<h2>Answer : Option A) true</h2><h3>Explanation : </h3>

CO which is carbon monoxide is a primary pollutant for air pollution. As it contributes to the major proportion in the air pollutants amongst others. It has 48% percent of pollutants in comparison to other pollutants of air. The source of carbon monoxide is mainly from incomplete combustion of fuel mostly from the older cars.

The rise in atmospheric CO results in fatigue, drowsiness and my also sometimes lead to death of the person exposed to it.

nexus9112 [7]3 years ago

3 0

Answer:

True

Explanation:

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3 years ago

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Anika [276]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Be sure to answer all parts.

sergey [27]

The moles of O2 are needed to react completely with 1.00 mol of C₂H₂ is 5 mol , 0.23 moles of C₂H₂ are needed to form 0.46 mol of CO₂ .

<h3>What is a Balanced Reaction ?</h3>

A balanced reaction is a reaction in which the number of atoms present in the reactants is equal to the number of atoms in the products.

It is given in the question that

the combustion of Acetylene is given by

2 H―C≡C―H + 5 O₂ → 4 CO₂ + 2 H₂O

Given moles of Acetylene = 1 mol

Moles of oxygen needed = 5 mol

as the mole fraction of Acetylene to Oxygen is 1 :5

The mole of Acetylene needed to form 0.46 mol of CO₂ is ?

mole fraction of CO₂ to Acetylene is 4:2

Therefore 0.23 moles of CO₂ is required.

To know more about Balanced Reaction

brainly.com/question/14280002

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5 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

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2 years ago