All categories

3 years ago

2C18H38 + 5502 → 36CO2 + 38H2O

Chemistry

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

Octadecane + Dioxygen = Carbon Dioxide + Water

Explanation:

Octadecane - C18H38

Oxidation Number

N-octadecane

Dioxygen - O2

Oxidation Number

O₂ Oxygen Gas Oxygen Molecular Oxygen Oxygen O-17

Products

Carbon Dioxide - CO2

Oxidation Number

Carbonic Acid Gas Dry Ice [CO2] Carbonic Anhydride R-744 E-290 E 290 E290

Water - H2O

Oxidation Number

Dihydrogen Monoxide Dihydridooxygen [OH2] HOH Hydrogen Hydroxide Dihydrogen Oxide Oxidane Hydrogen Oxide

You might be interested in

How many molecules are in 42.3g sample of water

Helga [31]

Answer:

The number of molecules is 1.4140*10^24 molecules

Explanation:

To know the number of molecules, we need to determine how many moles of water we have, water has molar mass of 18.015g/mol

This means that one mole of water molecules has a mass of 18.015g.

42.3g * 1 mole H2O/18.015g

= 2.3480 moles H2O

We are using avogadros number to find the number of molecules of water

2.3480 H2O * 6.022*10^ 23moles/ 1mole of H2O

That's 2.3480 multiplied by 6.022*10^23 divided by 1 mole of H2O

Number of molecules = 1.4140 *10^24 molecules

5 0

3 years ago

HURRY PLEASE

katrin2010 [14]

Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

Melting point is the temperature at which the substace changes its state from solid to liquid. During that change, the temperature of the substance does not change, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means that portion of the diagram is flat.

The same is valid during boiling: the temperature remains constant while the substance is passing from liquid to gas at the boiling point.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Between the melting (-183°C) and boiling (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a diagonal (slant) line between those points. During this interval the heat is used to increase the temperature and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

The table shows that the boiling point of gold is 2,856°C.

In a temperature-vs.-time diagram the temperature is represented on the vertical axis (y-value) and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during boiling, the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the top horizontal line in the diagram.

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

The table shows that the melting point of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the temperature - vs. - time diagrams show flat lines (constant y-values) during those intervals of time.

4 0

3 years ago

Read 2 more answers

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so

Illusion [34]

Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl

4 0

3 years ago

Read 2 more answers

How does carbon dioxide behave differently (its properties) as a gas compared with as a solid?

ankoles [38]

It is effected by diffusion (the power of smell and wind spread) but a solid is not.

4 0

3 years ago

___ Au₂S₃ + ___ H₂ → ___ Au + ___ H₂S

Natalija [7]

Answer:

2Au₂S₃ + 6H₂ → 4Au + 6H₂S

Explanation:

Balancing:

2Au₂S₃ + 6H₂ → 4Au + 6H₂S

6 0

3 years ago