All categories

2 years ago

How many moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned

Chemistry

1 answer:

ANTONII [103]2 years ago

8 0

Answer:

1.552 moles

Explanation:

First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.

This is illustrated below:

2C8H18 + 25O2 -> 16CO2 + 18H2O

Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 = 22.1g

Mole of C8H18 =..?

Number of mole = Mass /Molar Mass

Mole of C8H18 = 22.1/144

Mole of C8H18 = 0.194 mole

From the balanced equation above,

2 moles of C8H18 produced 16 moles of CO2.

Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.

Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned

You might be interested in

Q2. 0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht

Vera_Pavlovna [14]

Answer:

Mass concentration (g/L) = 2.49g/L.

Explanation:

No. of moles = $\frac{mass}{molar mass}$

= $\frac{0.254}{204}$ = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= $\frac{0.001245}{0.02}$ = 0.062 mol/L

We know that:

$C_{1} V_{1}$ = $C_{2} V_{2}$

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

∴Mass concentration (g/L) = 0.0498g ÷ 0.02L

= 2.49g/L.

3 0

2 years ago

The acid dissociation constant ka for an unknown acid ha is 4.57 x 10^-3 what is the base dissociation constant kb for th econju

SashulF [63]

Answer:

2.19 x 10^-12.

Explanation:-

The relation between Ka and Kb for an acid and it's conjugate base is

Ka x Kb = Kw where Kw = ionic product of water.

So Kb = 10^-14 / (4.57 x 10 ^ -3)

= 2.19 x 10^-12

4 0

2 years ago

What is the redox half equation for 3Ag2S + 2Al --> 6Ag + Al2, and identity which material is oxidized and which is reduced?

marta [7]

Answer:

Al is oxidized while Ag is reduced.

Explanation:

The complete molecular equation is;

3Ag2S + 2Al --> 6Ag + Al2S3

Oxidation half equation;

2Al ------> 2Al^3+ + 6e

Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

2Al + 6Ag^+ ----->2Al^3+ + 6Ag

Hence; Al is oxidized while Ag is reduced.

5 0

2 years ago

Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?

Marina CMI [18]

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

8 0

2 years ago

Which of the following is not true about compounds?

finlep [7]

Answer:

Elements can combine in any proportion to form a compound. -third choice

4 0

2 years ago

Read 2 more answers