What is the solution to the problem rounded to the correct number of significant figures? 8.01 × 4.1 = ? A. 32.8 B. 32.84 C. 32.
2 answers:
<u>Answer:</u> The correct option is Option D.
<u>Explanation:</u>
Significant figures are defined as the figures that represent the digits of a number which carry an important contribution to the numerical value, starting with the first non-zero digit. For Example: 105.268 has 6 significant figures because every digit carry its own contribution towards the numerical value.
Whenever there is multiplication, the answer will contain the same number of significant figures as there are in the least precise numerical value.
Significant figures in 8.01 are 3 and that in 4.1 are 2.
So, the answer of these two numerical value will contain 2 significant figures.
Hence, the correct option is Option D.
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Answer:
1s22s22p6: Neon (Ne)
1s22s22p63s23p3: Phosphorous (P)
1s22s22p63s23p64s1: Potassium (K)
1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)
1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4
Explanation:
you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al is required.
Therefore for 2 moles of Al,
Total F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
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