Ask question

Ask question

All categories

Vladimir79 [104]

3 years ago

11

3- Which process use to convert heavier hydrocarbons into lighter ( gasoline ) in the absence of oxygen and in the presence of c

atalyst ?
A- Fractional distillation

B- Melting

C- Cracking

D- Boiling

1 answer:

Ahat [919]3 years ago

7 0

C, catalytic cracking

You might be interested in

Solve and show work. Li2S + 2 HNO3 --> 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of

Elenna [48]

Li2S + 2 HNO3 --> 2 LiNO3 + H2S

Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S

Li S + H2 N2 O5 -> Li N2 O5 + H2 S

Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2

Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2

7 0

2 years ago

To describe the length of a classroom, a student should use

Kamila [148]

The answer is D).....

8 0

3 years ago

Read 2 more answers

Please help i will give brainliest

ExtremeBDS [4]

Question 2 answer is A

7 0

3 years ago

What is the molality of a solution when 0.50 moles glucose is dissolved in 750 g of water?

xxMikexx [17]

Answer:

0.67mol/Kg

Explanation:

The following were obtained from the question:

Mole of solute = 0.50mol

Mass of solvent = 750g = 750/1000 = 0.75Kg

Molality =?

Molality = mole of solute /mass of solvent

Molality = 0.5/0.75

Molality = 0.67mol/Kg

3 0

3 years ago

how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *

Vitek1552 [10]

Answer:

The number of moles =

$Moles=4.97\times 10^{-8}$

The number of molecules =

$Molecules = 2.99\times 10^{16}$

Explanation:

Volume of the sphere is given by :

$V=\frac{4}{3}\pi r^{3}$

here, r = radius of the sphere

$radius=\frac{diameter}{2}$

$radius=\frac{6.0}{2}$

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

$V=\frac{4}{3}\pi 0.03^{3}$

$V=\frac{4}{3}\pi 2.7\times 10^{-5}$

$V=1.13\times 10^{-4}dm^{3}$

$V=1.13\times 10^{-4}L$ (1 L = 1 dm3)

Now, concentration "C"=

$C=4.4\times 10^{-4}moles/liter$

The concentration is given by the formula :

$C=\frac{moles}{Volume(L)}$

This is also written as,

$Moles = C\times Volume$

$Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}$

$Moles=4.97\times 10^{-8}$ moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n" mole of substance contain =( n x Na )

$N_{a}=6.022\times 10^{23}$

Molecules =

$Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}$

$Molecules = 2.99\times 10^{16}$ molecules

7 0

3 years ago