HELPPP MEE I NEED HELPP PLEASE

The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO

borishaifa [10]

Answer:

710,33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 used to combust the octane

$87.54 mol O2 x \frac{15}{85}$ = 15.44 mol O2 used to form NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710,33 g NO2

4 0

3 years ago

Read 2 more answers

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

c)

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

d)

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

3 years ago

How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.345 g of Fe2S3 if the percent yiel

GuDViN [60]

Answer:

25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃

Explanation:

Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.

0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):

0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = 1.6595x10⁻³ moles Fe₂S₃

Moles of Fe needed to produce these moles of Fe₂S₃ are:

1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =

3.3189x10⁻³ moles of FeCl₃

As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:

3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = 5.106x10⁻³ moles of FeCl₃

A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:

5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L =

<h3>25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃</h3>

8 0

3 years ago

Which is an example of a heterogeneous mixture? A coffee B soil C gelatin D air

mariarad [96]

Heterogeneous mixtures are made of different substances that remain physically seperate. An example would be mixing sand and sugar together.
the answer would be soil
The answer is B soil

6 0

3 years ago

The concentration of hydrogen peroxide solution can be determined by

max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- = 0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution = 5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0

3 years ago