Using the periodic table, complete the table to describe each atom. Type in your answers.

The oxidation state numbercan aid in identifying the unknown element present in both compounds. They denote the number of electrons that are either donated, received, or shared to yield compounds.

Remember the fundamental principles governing oxidation numbers.

1. In a neutral compound, the total of all oxidation numbers is zero.
2. Chlorine, bromine, and iodine typically exhibit an oxidation number of -1(unless paired with fluorine and oxygen)

Assume the oxidation state for element Mis designated as x.
Referring to rule 2, chlorine possesses an oxidation state of -1.

Now, for the compound MCl₂ (which is neutral), the equation can be formulated as
x + (2 * -1)= 0 ⇒ x₁= +2

For MCl₃, the corresponding equation is
x + (3 * -1)=  0 ⇒ x₂= +3

This indicates that the elementhas two distinct oxidation statesin its compounds, which are +2and +3.

The identified element is iron (Fe), as it shows +2 and +3 oxidation states across these compounds.

Memorizing this is essential. Regrettably, there isn't a simpler method to tackle these oxidation states.

The final answer is iron (Fe).

1 percent of the solar radiation

Answer:

a)

b) 1657 €

Explanation:

Hola,

a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:

A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:

b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:

En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:

Finalmente, usando regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

Answer:

The empirical formula is =

The Valproic acid formula is =

Explanation:

Mass of the produced water = 0.166 g

Molar mass of water = 18 g/mol

The moles of are calculated as 0.166 g /18 g/mol = 0.00922 moles.

In 1 mole of water, there are 2 moles of hydrogen atoms.

Thus,

Moles of H = 2 x 0.00922 = 0.01844 moles

Each hydrogen atom's molar mass is 1.008 g/mol.

Hydrogen mass in the molecule = 0.01844 x 1.008 = 0.018588 g

Mass of produced carbon dioxide = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

The moles of are calculated as 0.403 g  /44.01 g/mol = 0.009157 moles.

Each carbon atom's presence is 1 mole in 1 mole of carbon dioxide.

So,

Moles of C = 0.009157 moles

The molar mass of carbon is 12.0107 g/mol.

Carbon mass in molecule = 0.009157 x 12.0107 = 0.11 g

Since Valproic acid comprises only hydrogen, oxygen, and carbon, the oxygen mass in the sample = Total mass - Carbon mass  - Hydrogen mass.

The sample's overall mass = 0.165 g.

Oxygen mass in the sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  

The molar mass of oxygen is 15.999 g/mol.

Moles of O  = 0.036412  / 15.999  = 0.002276 moles

Taking the simplest ratio for H, O, and C yields:

0.01844: 0.002276: 0.009157

= 8: 1: 4

The empirical formula becomes

While molecular formulas detail the precise count of atoms for each element, empirical formulas represent the simplest form or reduced ratio of these elements in the compound.

Consequently,  

The molecular mass equals n × Empirical mass.

Here, n is a positive integer from 1, 2, 3...

Empirical mass = 4×12 + 8×1 + 16 = 72 g/mol.

Molar mass = 144 g/mol.

Thus,  

The molecular mass = n × Empirical mass.

144 = n × 72

⇒ n = 2

The Valproic acid formula is =

Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

Answer: cesium