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3 years ago

What was the average speed of the truck for this trip?

Chemistry

1 answer:

Goshia [24]3 years ago

8 0

Answer:

31.67 mph

Explanation:

To calculate the average speed of the truck, we must first obtain the total distance travelled by the truck followed by the total time taken for the truck to cover the distance travelled.

The following data were obtained from the question include:

Total distance) = 30 + 45 + 50 + 65 = 190 miles

Total time = 1 + 2 +1 +2 = 6 hours

Average speed =.?

Average speed = Total distance / Total time

Average speed = 190 /6

Average speed = 31.67 mph

Therefore, the average speed of the truck is 31.67 mph

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Answer : The rate of reaction is,

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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 :

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

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

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

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,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

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$Rate=k[NO_2][O_3]$

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Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

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$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

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