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anygoal [31]
3 years ago
12

An atom bonds with another atom. What is the best classification for this reaction?

Chemistry
2 answers:
defon3 years ago
5 0
Chemical would be the correct answer
kiruha [24]3 years ago
3 0

The best classification of reaction in which  an  atom bonds with another atom   is <u>chemical reaction</u>

<em><u> </u></em><u><em>Explanation</em></u>


In a chemical reaction, reactants come into contact to  each other.

Bond   between  atoms in the reactant  are broken down and then rearrange to form new  bonds  which form product


For   example of a chemical reaction is as below

N2 + 3H2 → 2NH3

N2 and H2  are atoms that come in to contact to each  other to form NH3.

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How is a mixture such as a plate of pasta different from a mixture such as air
stealth61 [152]

Answer:

Well there is a lot of differences between the two. Its called homogeneous and Heterogeneous mixtures. Homogeneous mixtures are all the substances are evenly distributed throughout the mixture (salt water, air, blood).  Heterogeneous mixtures are the substances that are not evenly distributed (chocolate chip cookies, pizza, rocks). So Pasta with sauce and meatballs is heterogeneous and air is homogeneous




HOPE THIS HELPS HAVE A GREAT DAY!!~

Explanation:

6 0
2 years ago
which of the following requires the most energy to break a bond? a. breaking cl-br bond. b. breaking a n-p bond. c. breaking a o
irina1246 [14]

A. Breaking a Cl-Br bond

6 0
2 years ago
HELP ASAP PLEASAE!!!
N76 [4]

I believe the answer is A. However, I would double check the formula.

4 0
2 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
Oxidation of Nitrogen in NO2​
tiny-mole [99]

Explanation:

oxidation of Nitrogen in NO2 is +4

8 0
2 years ago
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