Chromatography and fractional distillation can be used to separate compounds.

Consider the positions of barium (Ba), sulfur (S), silicon (Si), and calcium (Ca) on the periodic table. The atoms of which elem

Rudiy27

Element atomic number    position

Ba 56    group 2, period 6

Ca 12    group 2, period 3

S 16 group 16, period 3

Si `14    group 14, period 3

Now, you need to know the properties of the different type of elements and the tendencies on the periodic table.

The metallic elements are, those placed on the left side of the periodic table, are the ones that release an electron more easily, so they will requiere less energy to give it up when forming chemical bonds.

The higher the metallic character the less the energy need to give up an electron.

The metallic character grows as the group number decreases (goes to the left) period increases (goes downward), so among the elements considered, Barium will require the least amount of energy to give un an electron when forming chemical bonds.

5 0

3 years ago

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tres pensamientos negativos sobre el C0Vid-19 que se vive en la actualidad y luego transformemos a pensamientos positivos.

katrin2010 [14]

Answer:

Bueno,

Explanation:

Tres pesnamientos negativos sobre el C0Vid-19 son,

- Encerrados en la casa

- Tener que ponerse mascara

- Todos los lugares divertidos cerrados

Convertidos en positivos:

- Pasamos mas tiempo de calidad en familia.

- Nos protejemos unos a otros.

- Descansan un poco todas las personas que se han pasado toda la vida trabajando.

7 0

3 years ago

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What is the square root of 25

Sauron [17]

Answer:

5

Explanation:

because 5x5 = 25

and 25 ÷5= 5

4 0

3 years ago

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20. Tell one thing that is true about these balanced chemical equations. Explain your answer. A + B → CDE. H2 + O2 → H2O2.

sweet-ann [11.9K]

D because the mass evens out the objct

4 0

3 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

Answer:

For a: The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b: The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c: The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a:

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b:

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c:

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

3 0

3 years ago