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3 years ago

(3 Points)

Chemistry

2 answers:

8090 [49]3 years ago

8 0

Answer:The correct answer is option C.

Explanation:

Radio waves and microwaves both are the examples of electromagnetic waves.

Radio waves: These are waves with lowest frequency in the electromagnetic spectrum with longest wavelength.

Microwaves : These are the second lowest frequency waves in the electromagnetic spectrum with wavelength shorter than radio waves.

$E=h\nu$

E = Energy of an electromagnetic wave

$\nu=\frac{c}{\lambda }$

$\nu \propto \frac{1}{\lambda }$

$E\propto \frac{1}{\lambda }$

$\nu$ = Frequency of the electromagnetic wave

$\lambda$ = Wavelength of the electromagnetic wave

Radio waves with longest wavelength has lower radiant energy than the microwaves.Hence, correct option is C.

prisoha [69]3 years ago

5 0

Your answer is B. radio waves have shorter wavelenghts than microwaves.
Have a great day

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3 years ago

Write the formula for potassium oxide. why do you not need prefixes in the name

Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

Potassium- K- on the left end of period four.
Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

Potassium is a metal,
Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.

Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion $\text{O}^{2-}$ when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each $\text{O}^{2-}$ ion is twice that on a $\text{K}^{+}$ ion. Each $\text{K}^{+}$ would pair up with two $\text{O}^{2-}$ . Hence the subscript in the formula: $\text{K}_{\bf 2}\text{O}$ .

There are two classes of compounds:

Covalent compounds, which need prefixes, and
Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide $\text{CO}_2$ , the prefix di- indicates that there are two oxygen atoms in the formula $\text{CO}_2$ . However, there's no need for prefix in ionic compounds such as $\text{K}_2\text{O}$ .

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3 years ago

What is the expected oxidation state for the most common ion of element 2

lozanna [386]

Answer: 1+

Justification:

The ionization energies tell the amount of energy needed to release an electron and form a ion. The first ionization energy if to loose one electron and form the ion with oxidation state 1+, the second ionization energy is the energy to loose a second electron and form the ion with oxidation state 2+, the third ionization energy is the energy to loose a third electron and form the ion with oxidation state 3+.

The low first ionization energy of element 2 shows it will lose an electron relatively easily to form the ion with oxidations state 1+.

The relatively high second ionization energy (and third too) shows that it is very difficult for this atom to loose a second electron, so it will not form an ions with oxidation state 2+. Furthermore, given the relatively high second and third ionization energies, you should think that the oxidation states 2+ and 3+ for element 2 never occurs.

Therefore, the expected oxidation state for the most common ion of element 2 is 1+.

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