The electronic configuration of Aluminum is 2,8,3.
<h3><u>Explanation:</u></h3>
Aluminum is a metallic element belonging to group 3 of periodic table. Its atomic number is 13.
The rules of electronic configuration is
A. The maximum number of electrons a shell can hold is determined by 2n² where n is the shell number numbered from the one as 1 which is nearest to the nucleus.
B. Each shell has some subshells which actually holds the electrons. Shell 1 has only s subshell where as shell 2 has s and p subshells and so on.
C. Afbau's principle states which subshell of which shell is first filled. For example, 4s subshell is filled before 3d subshell which is stated by Afbau's principle.
The atomic number 13 is filled like- 1st shell has 2 electrons, 2nd shell has 8 electrons and 3rd shell has 3 electrons.
It is also represented as 1s², 2s²,2p⁶, 3s² 3p¹.
Answer:
4.81×10¹⁰ atoms.
Explanation:
We'll begin by converting 3.2 pg to Ca to grams (g). This can be obtained as follow:
1 pg = 1×10¯¹² g
Therefore,
3.2 pg = 3.2 pg × 1×10¯¹² g / 1 pg
3.2 pg = 3.2×10¯¹² g
Therefore, 3.2 pg is equivalent to 3.2×10¯¹² g
Next, we shall determine the number of mole in 3.2×10¯¹² g of Ca. This can be obtained as follow:
Mass of Ca = 3.2×10¯¹² g
Molar mass of Ca = 40.08 g/mol
Mole of ca=.?
Mole = mass /molar mass
Mole of Ca = 3.2×10¯¹² / 40.08
Mole of Ca = 7.98×10¯¹⁴ mole.
Finally, we shall determine the number of atoms present in 7.98×10¯¹⁴ mole of Ca. This can be obtained as illustrated below:
From Avogadro's hypothesis,
1 mole of Ca contains 6.02×10²³ atoms.
Therefore, 7.98×10¯¹⁴ mole of Ca will contain = 7.98×10¯¹⁴ × 6.02×10²³ = 4.81×10¹⁰ atoms.
Therefore, 3.2 pg of Ca contains 4.81×10¹⁰ atoms.
<u>The </u><u>emperical formula </u><u>can be </u><u>calculated </u><u>as follows:</u>
<u>mass :</u> 4.12g of N and 0.88g of H
<u>moles:</u> 4.12g/14.007 and 0.88g/1.008
<u>moles:</u> 0.294 and 0.873
<u>moles/smallest number</u>: (gives the ratio of atoms)
0.294/0.294 and 0.873/0.294
1 and 2.969
1 and 3
The simplest ratio of N : H is 1 : 3
<u>Hence the </u><u>emperical formula </u><u>is </u><u>NH3</u>
<u>Explanation:</u>
1. The emperical formula can be found by getting the <u>simplest ratio of atoms in a molecule.</u>
2. To get the simplest ratio of atoms we find out the <u>number of moles of each atoms in it.</u>
3. From the number of moles of atoms we can find the simplest ratio of each atom by dividing the <u>number of moles by the smallest number.</u>
The molar mass of the gas : 18 x 10⁻³ kg/mol
<h3>Further explanation</h3>
Given
An unknown gas has one third the root mean square speed of H2 at 300 K
Required
the molar mass of the gas
Solution
Average velocities of gases can be expressed as root-mean-square (V rms)
T = temperature, Mm = molar mass of the gas particles , kg/mol
R = gas constant 8,314 J / mol K
v rms An unknown gas = 1/3 v rms H₂
v rms H₂ :
V rms of unknown gas =