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3 years ago

Is fireworks exploding a chemical change?

Chemistry

2 answers:

anygoal [31]3 years ago

8 0

Answer:

The explosion of fireworks is an example of chemical change. During a chemical change, substances are changed into different substances. In other words, the composition of the substance changes.

zalisa [80]3 years ago

7 0

Answer:

Yes

Explanation:

The fireworks explode

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Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ

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Answer:

a. True

b. False

c. True

d. False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

$\frac{Products}{reagents}$

In any case

aA +Bb ⇌ Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

$Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}$

Using the ideal gas law, we can get a relationship between K and Kp

Pv=nRT where $P=\frac{n}{v}*RT$ we know that $\frac{n}{v}$ is the molar concentration. When we replace P in the expression for Kp we get:

$Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}$

Reorganizing the equation:

$Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}$

We can see K in the expression

$Kp= K*(RT)^{c+d-a-b}$

Delta n = c+d-a-b

$Kp= K*(RT)^{delta n}$

For the reaction

$H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)$

Delta n = 2-1-1=0

$Kp= K*(RT)^{0}$

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

$CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)$

$K= [CO_{2}]$

So we can see that $[CaCO_{3}]$ is not include in the expression.

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)

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