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Varvara68 [4.7K]
3 years ago
13

What does the subscript 2 indicate in the formula o2?

Chemistry
1 answer:
Tpy6a [65]3 years ago
6 0

The subscript 2 in the formula o2 indicates that there are 2 atoms of the element oxygen, represented by the symbol "O."   The subscript to the right of the symbol tells how many atoms are present of that element.  

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For a principal quantum number n, how many atomic orbitals are possible?
slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

5 0
3 years ago
What property is primarily responsible for determining the type of electromagnetic energy and peak wavelength emitted by a star
svetoff [14.1K]
The answer i think it is... is Temperature 
Hope this helped!
6 0
4 years ago
Read 2 more answers
Describe a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2)
Lady bird [3.3K]

A qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq)  →   BaSO₄(s).

<h3>What is qualitative test?</h3>

Qualitative test measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation of the sample.

<h3>Qualitative test for sulfate in alum crystals </h3>

When an aqueous solution of a barium salt (BaCl₂) is mixed with an aqueous solution containing sulfate, a white precipitate of insoluble BaSO₄ forms according to the net ionic equation given below;

Ba²⁺(aq) + SO₄²⁻ (aq)  →   BaSO₄(s)

Thus, a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq)  →   BaSO₄(s).

Learn more about qualitative test here: brainly.com/question/2109763

#SPJ1

8 0
2 years ago
What is the isotope notation for an ion of silver-109 with a charge of positive 1
Arisa [49]

Answer:

The isotope notation for an ion of silver-109 with a charge of positive 1 is _{47}^{109}Ag^{1+}.

Explanation:

The elements with same atomic number with different mass numbers are called isotopes.

The isotopes of silver is as follows.

_{47}^{109}Ag,_{47}^{107}Ag

From the above two isotopes have same atomic number and different mass number.

From the given,the isotope notation for an ion of silver-109 with a charge of positive 1

It can be represented is as follows.

_{47}^{109}Ag^{1+}

General representation of isotope is as in attachment.

6 0
3 years ago
Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo
lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, n_i=4

Final orbit of electrons, n_f=2

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

Putting all the values we get :

E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV

We know that : 1\ eV=1.6\times 10^{-19}\ J

So,

E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J

Energy of wave in terms of frequency is given by :

E=hf

f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz

Also, c=f\lambda

\lambda is wavelength

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm

Hence, this is the required solution.

4 0
3 years ago
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