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OleMash [197]
2 years ago
5

Why are there different ways to describe an atom? What are the ways discussed in the reading, and what are the benefits and draw

backs of each?
Chemistry
1 answer:
NISA [10]2 years ago
8 0

Answer:

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. Each electron is negatively charged. The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons.

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1 Condensation
larisa86 [58]

Answer:

Explanation:

A ) Solar Radiation

B ) Evaporation

C ) Transpiration

D ) Condensation

E ) Precipitation

F ) Run Off

G ) Infiltration

H ) Groundwater .

3 0
3 years ago
Given the following reaction and data, A + B → Products
natima [27]

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

<h3>Rate = k×[A]</h3>

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

<h3>k = 0.213s⁻¹</h3>

6 0
3 years ago
Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión
miv72 [106K]

Answer:

Las siguientes son reacciones químicas;

combustión de leña

oxidación del hierro

descomposición del agua en hidrógeno y oxígeno

Explanation:

Una reacción química da como resultado la formación de una (s) sustancia (s) nueva (s), mientras que un cambio físico no conduce a la formación de una sustancia nueva.

Las siguientes son reacciones químicas;

combustión de leña: la combustión de madera implica la oxidación del carbono según la reacción; C (s) + O2 (g) -------> CO2 (g)

oxidación del hierro: La oxidación del hierro conduce a la formación de óxidos de hierro. Como; 2Fe (s) + O2 (g) ----> 2FeO (s)

descomposición del agua en hidrógeno y oxígeno: esta es una reacción química en la que el agua se descompone de la siguiente manera; 2H2O (l) -----> 2H2 (g) + O2 (g)

Todos estos procesos enumerados anteriormente conducen a la formación de nuevas sustancias, por lo tanto, son reacciones químicas.

3 0
2 years ago
Consider the electron configuration. mc011-1.jpg mc011-2.jpg mc011-3.jpgmc011-4.jpgmc011-5.jpg mc011-6.jpg mc011-7.jpgmc011-8.jp
zepelin [54]
The element which has the electronic configuration is CHLORINE.
The atomic number of chlorine is 17 and it has 7 valence electrons in its outermost shell. Because it needs only one more electrons to have a stable octet, it usually react with metals from group one of the periodic table who are normally willing to donate the single electrons in their outermost shells. The ground state electronic configuration of chlorine atom is 1S^2 2S^2 2P^6 3S^2 3P^5.
5 0
3 years ago
Read 2 more answers
29.47 mL of a solution of the acid HBr is titrated, and 72.90 mL of 0.2500-M NaOH is required to reach the equivalence point. Ca
Klio2033 [76]

The original concentration of the acid solution is 6.175 \times 10^-4 mol / L.

<u>Explanation:</u>

Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume

                        HBr + NaOH -----> NaBr + H2O

There is a 1:1 equivalence with acid and base.

Moles of NaOH = 72.90 \times 10^-3 \times 0.25

                          = 0.0182 mol.

[ HBr ] = moles of base / volume of a solution

          = 0.0182 / 29.47

          = 6.175 \times 10^-4 mol / L.

4 0
3 years ago
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