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____ [38]
3 years ago
9

C(graphite) + O2(g) → CO2(g) + 94.05 kcal C(diamond) + O2(g) → CO2(g) + 94.50 kcal What could you infer from the information giv

en above?
Chemistry
2 answers:
olga nikolaevna [1]3 years ago
7 0
Looking at both reactions, we can see that the combustion of carbon, graphite or diamond, leads to the formation of carbon dioxide and energy. Since energy is a product of the reaction, we known that this is an exothermic reaction. The value of the change in enthalpy, ΔH, will be negative.

A negative value of ΔH in each exothermic reaction suggests that the carbon dioxide product is at a lower energy than the reactants. And since more energy is released in the combustion of diamond compared to graphite, we know that diamond has a higher internal energy than graphite.
mr_godi [17]3 years ago
7 0

Answer: E 2 - E 1 is negative. and

Diamond has a higher enthalpy than graphite.

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How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?
Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g)  + 5O2(g)  ----> 3CO2(g)   + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0


7 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST TO PERSON WHO ANSWERS FASTEST AND CORRECT BUT ANY ANSWERS TO JUST GET POINTS WILL GET REPORTED AND DELETED
AleksandrR [38]

Answer:

B

Explanation:

When water is at the surface of the ground, it evaporates and goes back to the clouds for which causes it to rain therefore returning water back.

3 0
3 years ago
A student measures the pressure and volume of an empty water bottle to be 1. 4 atm and 2. 3 l
rewona [7]

The new volume of the bottle is 5.07 L.

<h3>What is Boyle's Law?</h3>

It is a gas law that states the pressure decrease with the increase in the pressure.

By the formula

\rm P_1V_1= P_1V_2

Given that, volume1 is 2.3l

Pressure1 is 4 atm.

Pressure2 to 0.65 atm.

V2 is to find?

Putting the values in the equation

\rm 1.4\times 2.3 = 0.65 \times V_2\\\\V_2 = 5.07 L

Thus, the new volume of the bottle is 5.07 L.

Learn more about Boyle's Law

brainly.com/question/1437490

#SPJ4

6 0
2 years ago
What's ligand and how are they classified​
KATRIN_1 [288]

Explanation:

<u>Ligands:</u> In co-ordination chemistry ligands are ion, molecule or any species which donates electron pair to central metal atom.

Depending on the type of interaction Ligands are of three types.

  1. Sigma donor only
  2. sigma as well as pi donor
  3. pi acceptor ligand

let's understand each type of Ligands individually & in more detail.

1 - Sigma donor only: This is a unidirectional interaction, in which filled ligand overlaps (head to head) with central metal atom/ion & donates pair of electron in the LUMO of metal.

generally all the molecules of 2nd period without pi bond comes in this category, below are few example of sigma donor ligands,

\small \sf NH_3, H_2O, CH_3^-, H^-, R-OH, R-NH_3, etc

2- Pi donor: This in also a unidirectional interaction between ligand & central metal atom but the along with head to head overlap, side overlapping takes place.

generally protonated neutral molecules who have more than one pair to donate show such interaction, for e.g.

NH3 have two lone pair to donate but the energy level of both the lone pairs are different hence when it is neutral it only donates one pair of electron. but when NH3 is protonated to NH2- it have two electron pairs (negative charge+ lone pair) to donate & both the pairs have same energy level. example of such ligands are below,

\sf \small NH_2^-, OH^-, R-O^-, R-NH^-, F^-, Cl^-, Br^- SH^- etc

3- Pi acceptor ligand: This is a bidirectional interaction between ligand & central metal atom/ion, the filled orbital of ligand undergoes head to head to overlap with vacant orbital of central metal atom, & filled D orbital of central metal donates their pair to vacant LUMO of ligand.

depending on the LUMO pi acceptor ligands are further classified into two categories.

d\pi - \sigma*   \small \sf When  \: lumo \:  is  \: \sigma*\\ d\pi - \pi*   \small \: \sf When  \: lumo  \: is  \: \pi*

The dπ-σ* is seen in molecules of 3rd period onwards without pi bond <em>for e.g.</em>

<em>PH3,</em><em> </em><em>PR</em><em>3</em><em>,</em><em> </em><em>AsR</em><em>3</em><em> </em><em>&</em><em> </em><em>SR</em><em>2</em><em> </em><em>etc</em>

The dπ-π* is seen in molecules of 2nd or3rd period with pi bond <em>for e.g.</em>

CO C N- SC N^- etc

<em><u>Thanks for joining brainly community!</u></em>

8 0
2 years ago
How conditions on Earth would be different if its axis of rotation were vertical instead of tilted.
kykrilka [37]

Answer:

there would be different times because the sun and moon are in different places and one side would be sunny for a long time

Explanation:

3 0
3 years ago
Read 2 more answers
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