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3 years ago

Which stage in the free radical substitution of methane by chlorine will have the lowest activation

1 answer:

6 0

I am pretty sure that the stage in the free radical substitution of methane by chlorine which will have the lowest activation energy is B Cl • + Cl • → Cl 2. I choose this option because during termination there is no breaking of bonds within two reactive radicals which are joined together.

Do hope it will help you!

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

In bullet points: explain the difference between thermal energy, heat and temperature.

sergeinik [125]

Thermal Energy- is the energy that comes from heat. This heat is generated by the movement of tiny particles within an object. The faster these particles move, the more heat is generated.

Heat-energy is the result of the movement of tiny particles called atoms, molecules or ions in solids, liquids and gases. Heat energy can be transferred from one object to another. The transfer or flow due to the difference in temperature between the two objects.

Temperature- is a measure of how hot or cold something is; specifically, a measure of the average kinetic energy of the particles in an object, which is a type of energy associated with motion.

3 0

3 years ago

How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?

Nata [24]

Answer:

$\boxed{\text{53 g }}$

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r: 142.04

2NaOH + … ⟶ Na₂SO₄ + …

n/mol: 0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces $\boxed{\text{53 g }}$ of Na₂SO₄.

4 0

3 years ago

A chemist requires 0.912 mol Na2CO3 for a reaction. How many grams does this correspond to?

mina [271]

0.912 mol Na2CO3 × 105.9888g Na2CO3/ 1 mol Na2CO3 = 96.66 grams

3 0

3 years ago

Which would be the biggest long-term benefit of building a city recycling center? A. building the new recycling building

igor_vitrenko [27]

C. reducing the amount of garbage

Explanation:

The biggest long-term benefit of building a city's recycling building is that it helps to reduce that amount of garbage.

Recycling is simply the collection and processing of waste materials into new products that can be used again.

Waste recycling helps to reduce that actual proportion and percentage of materials to be considered waste.
The overall long term goal of any recycling procedure is to reduce the amount of garbage in the environment.

learn more:

Untreated wastes brainly.com/question/4080706

#learnwithBrainly

7 0

3 years ago