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2 years ago

Benzene is the simplest example of _____ compound.

1 answer:

4 0

Benzene is the simplest example of the aromatic compound
Firstly, we want to explain what is the meaning of aromaticity?
the aromaticity is called on any compound which has cyclic shape and has a resonance. Resonance is delocalization of the electrons (which form the double bond) in the molecules, in other words, the double bond moves through the compound by the movement of electrons which leads to increasing the stability of the compound by this process which called resonance. The resonance may occur in open or cyclic compounds, and if it occurs in cyclic compounds, these compounds are called aromatic compounds. So, the benzene considered to be the simplest aromatic compound where it is not only has a resonance in a cyclic ring but also it was the first aromatic compound discovered and also, the most common aromatic compounds are derivatives of it (benzene).

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If an atom should form from its constituent particles,

tia_tia [17]

Answer:

(b) matter is lost and energy is released

Explanation:

When atoms are being formed from its constituent components it weighs less this is called mass defect so the answer would be (b) matter is lost and energy is released.

8 0

2 years ago

Question attached ! asap please

bezimeni [28]

Answer:

A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.

b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules

c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.

d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.

e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids

Explanation:

6 0

2 years ago

What the requirement for a reversible reaction to be at equilibrium ?

lana66690 [7]

A reversible reaction is a chemical change in which the products can be converted back to the original reactants under suitable conditions. In a reversible reaction, changing the reaction conditions e.g. concentration, pressure or temperature will change the net direction the reaction goes i.e. more to the right (forward) or more to left (backward).It also means a reversible reaction does not go to completion in either direction and all components, original reactants or ensuing products, ALL co-exist in the reaction mixture (see notes on chemical equilibrium).This means the reaction can go in either direction i.e. A + B ==> C + D or C + D ==> A + BA reversible reaction is shown by the sign , a half-arrow to the right (direction of forward reaction), and a half-arrow to the left (direction of backward reaction).It is really important you understand that the terms right & left AND forward & backward are used in the context of how the equation is presented.Most reactions are not reversible (irreversible) and have the usual complete arrow only pointing to the right.

4 0

3 years ago

Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2

Svetlanka [38]

Answer: The equilibrium concentration of $H_2(g)$ at 700 degrees Celsius is 0.0012 M

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c
$

Moles of $H_2S$ = 0.29 mole

Volume of solution = 3.0 L

Initial concentration of $H_2S$ = $\frac{0.29mol}{3.0L}=0.097M$

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial conc. 0.097 M 0M 0M

At eqm. conc. (0.097-2x) M (2x) M (x) M

The expression for $K_c$ is written as:

$K_c=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$K_c=\frac{(2x)^2\times x}{(0.097-2x)^2}$

$9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.097-2x)^2}$

$x=0.00060$

Equilibrium concentration of $[H_2]$ = 2x= $2\times 0.00060=0.0012M$

6 0

2 years ago

A mole of X reacts at a constant pressure of 43.0 atm via the reaction

liberstina [14]

Below is the solution:

-62.0 kJ ΔH = -75.0 kJ, P = 43.0 atm 43atm x 101325 Pa/1 atm = 4356975 Pa or N/m2 ΔV = 2.0 L - 5.0 L = 3.0 L = -3.0 x 10-3 m3 w = -PΔV = -(4356975 N/m2) (-3.0 x 10-3 m3) = 13070 N m = 13070 J = 13.07 kJ At constant P, q = ΔH = -75.0 kJ & ΔE = q + w = - 75.0 kJ + 13.07 kJ = -61.93 kJ = -62.0 kJ

5 0

3 years ago

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