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4 years ago

Apart from cases of nuclear processes (including fission), elements cannot be broken down to form other substances.

Chemistry

1 answer:

Shtirlitz [24]4 years ago

7 0

Explanation:

Element is ur answer

hope it helps u

plz mark as brainlist

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In the following pair, determine whether the two represent resonance contributors of a single species or depict different substa

Aleksandr-060686 [28]

Answer:

They are resonance contributors

Explanation:

Resonance structures are structures that differ only in the distribution or placement of electrons.

Considering the two structures, we can easily see that the two species have the same total number of bonds and electrons differing only in the distribution of these electrons.

Hence, they are resonance contributors.

5 0

3 years ago

The building block of all matter is a) water b) oxygen c) an atom d) an element

jeka94

My best try would be d an element

4 0

3 years ago

I have a few questions from my homework and I was wondering if I can get some help

sashaice [31]

Well it breaks down into small parts

3 0

3 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

What does the second law of thermodynamics say about entropy?

Annette [7]

Answer:

the ansnwer is c mf kf n dsj njnr kgjrk nj ee h ewfkjhnf ewjn fnkwfnjnfnfunf

Explanation:

because is sooo goood

8 0

3 years ago

Read 2 more answers