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3 years ago

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4. Both the alkene C4H8 and the alkyne C4H6 have rigid bond structures. However, C4H8 can form three isomers, whereas C4H6 can f

orm only two isomers. Why is this so?

1 answer:

Murrr4er [49]3 years ago

3 0

Both alkene C4H8 and alkyne C4H6 have rigid bond structures. C4H8 can form three isomers while C4H6 can only form two isomers. Isomers are defined as two or more compounds having the same formula but with different molecular arrangements. The alkene contains 1 double bond which connects 2 carbons, so the other 4 hydrogen atoms can be rearranged. The alkyne only has 2 extra hydrogen atoms.

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Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

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Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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3 years ago

a sample of unknown material weighs 500 n in air and 200 n when immesersed in alcholol with a specfic gravity of 0.7 what is the

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Answer: The mass density is 1166.36 $kg/m^{3}$ .

Explanation:

Given: Weight of sample in air $(F_{air})$ = 500 N

Weight of sample in alcohol $(F_{alc})$ = 200 N

Specific gravity = 0.7 = $0.7 \times 1000 = 700 kg/m^{3}$

Formula used to calculate Buoyant force is as follows.

$F_{B} = F_{air} - F_{alc}\\= 500 - 200 \\= 300 N$

Hence, volume of the material is calculated as follows.

$V = \frac{F_{B}}{\rho \times g}$

where,

$F_{B}$ = Buoyant force

$\rho$ = specific gravity

g = acceleration due to gravity = 9.81

Substitute the values into above formula.

$V = \frac{F_{B}}{\rho \times g}\\= \frac{300}{700 \times 9.81}\\= \frac{300}{6867}\\= 0.0437 m^{3}$

Now, mass of the material is calculated as follows.

$mass = \frac{F_{air}}{g}\\= \frac{500 N}{9.81}\\= 50.97 kg$

Therefore, density of the material or mass density is as follows.

$Density = \frac{mass}{volume}\\= \frac{50.97 kg}{0.0437 m^{3}}\\= 1166.36 kg/m^{3}$

Thus, we can conclude that the mass density is 1166.36 $kg/m^{3}$ .

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3 years ago