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tamaranim1 [39]
4 years ago
9

Which type of reaction does this diagram represent?

Chemistry
2 answers:
Mice21 [21]4 years ago
8 0
I want to say addition. But I have a tendency to be wrong

tensa zangetsu [6.8K]4 years ago
7 0

Answer: addition

Explanation: Addition: A reaction is one in which a molecule adds to another molecule to form a single product.

Condensation: A reaction in which molecules combine to form products along with the formation of a small molecule called as byproduct.

Elimination: A reaction in which small molecules are eliminated when a certain substance is made to react.

Substitution: A reaction in which one substance replaces another substance.

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In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
Calculate the volume 3.00 moles of a gas will occupy at 24.0˚C and 1.003 atm. *
vitfil [10]

Answer: 72.93 litres

Explanation:

Given that:

Volume of gas (V) = ?

Temperature (T) = 24.0°C

Convert 24.0°C to Kelvin by adding 273

(24.0°C + 273 = 297K)

Pressure (P) = 1.003 atm

Number of moles (n) = 3 moles

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K

1.003 atm•V = 73.15 atm•L

Divide both sides by 1.003 atm

1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm

V = 72.93 L

Thus, the volume of the gas is 72.93 litres

5 0
3 years ago
Organic Chem Rxn Question
NemiM [27]

Answer:

a, g, c

Explanation:

The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.

The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.

8 0
3 years ago
True/False: All of the elements combine in a variety of ways to make up all living and all nonliving things
likoan [24]

Answer:

true

Explanation:

5 0
3 years ago
In which species does phosphorous have the highest oxidation number?
nignag [31]

the compounds in which phosphorous posses the highest possible oxidation have to mention here.

The species in which phosphorous have the highest oxidation state are: H₃PO₄, P₂O₅, PCl₅

The possible oxidation state of phosphorous is III and V. The highest oxidation state is V. There are several compounds in which phosphorous posses the +5 oxidation state. Like- Phosphoric acid (H₃PO₄), phosphorous pentoxide (P₂O₅), Phosphorous chloride (PCl₅) etc.

The oxidation state of an element depends upon the valence electron the valence shell of phosphorous is 3s² 3p³. Thus there are 5 electrons, as it has vacant 3d orbital thus it can easily form compound having +5 oxidation state.

7 0
3 years ago
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