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3 years ago

In poor quality fireworks what will you notice and why

2 answers:

8 0

The best answer I could find was when you Google it, that the fuse is of poor quality. I cannot leave you a link, but you can find it for yourself. Put in poor quality fireworks and all sorts of things will pop up. No pun intended.

Kisachek [45]3 years ago

7 0

Answer:

Explanation:

Hello!

Low quality fireworks have loosely bound mortars, shells and cannon fuses. Moreover, there are mighty chances that the cannon fuse or the visco fuse of the firework would be inappropriately small and have loosely wrapped protective layers. A good quality firework would always have a long cannon fuse with minimum of three protective layers wrapped around it, tightly. The three layers are, invariably, a combination of usual and lacquer coated wires. While the usual wires provide mechanical strength, the lacquer coated ones provide water resistance to the fuse and keep the fuse from falling apart.

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3. What are the energy levels of electrons?

KiRa [710]

Energy levels inside an atom are the specific energies that electrons can have when occupying specific orbitals.

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3 years ago

The overall reaction of ozone reacting to form oxygen has been proposed to occur in a reaction mechanism of: What is the role of

LekaFEV [45]

Answer:

3) O(g) is an intermediate; 2O3(g)→3O2(g)

Explanation:

The decomposition of ozone to yield oxygen occurs in a sequence of steps. The various non-elementary reactions involved constitute the reaction mechanism. In the sequence of reaction steps O(g) serves as an intermediate.

The overall reaction involves the conversion of two moles of ozone to three moles of oxygen as shown in the answer. Thus the O(g) is merely a reaction intermediate.

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3 years ago

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N

zhenek [66]

Answer : The value of equilibrium constant (K) is, $5.71\times 10^4$

Explanation :

First we have to calculate the concentration of $N_2,O_2\text{ and }N_2O$

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M$

and,

$\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

The expression for equilibrium constant is:

$K=\frac{[N_2O]^2}{[N_2][O_2]}$

Now put all the given values in this expression, we get:

$K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}$

$K=5.71\times 10^4$

Thus, the value of equilibrium constant (K) is, $5.71\times 10^4$

7 0

3 years ago

PLS HELP ME IN THIS QUESTION

Gnesinka [82]

Answer:

the first one is just water and the second one is gas

Explanation:

7 0

3 years ago

Read 2 more answers

Why doesn't Magnesium react with Magnesium Nitrate?

lesya [120]

When a metal reacts with a salt, it displaces the metal present in the salt only if the metal has higher relative reactivity.
in this case , the displacing metal is same as the metal present in salt. both (being same metal)have same relative reactivity. hence, no displacement . NO reaction.

3 0

3 years ago