All categories

2 years ago

Which subatomic particles exist in orbital energy levels, or shells?

Chemistry

2 answers:

Ann [662]2 years ago

7 0

Answer: Electrons

Explanation: As given by Schrodinger that the probability of finding an electron in the given space is known as Orbital.

Thus the subatomic particles that exist in orbital energy levels or shells are electrons.

Electron has an charge of a negative unit. It was discovered by the scientist J.J.Thomson.

Atom consist of protons and neutrons that are present in the nucleus . Electrons revolve around the nucleus.

Naily [24]2 years ago

6 0

The correct answer among the choices given is the first option. Electron is the subatomic particle that exist in a orbital energy levels or shells. At the first energy level, the only orbital available for the electrons is the 1s while in other orbitals there are more than one type of orbital.

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What is the predominant intermolecular force in CBr4

vovikov84 [41]

CBr4 is a symmetric tetrahedral molecule so it will be non-polar.

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3 years ago

It has more protons than Cl but less than K

timama [110]

Answer: Argon (Ar), which has 18 protons.

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3 years ago

The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

2 years ago

Read 2 more answers

How many moles of NaOH are in 13.25 mL of 0.323 M NaOH?

adell [148]

M=mol/L, 0.323M=mol/0.01325. Rework to solve for mol and bam! (I.e. times the two numbers)

7 0

3 years ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

8 0

2 years ago

Read 2 more answers