Answer:
Iron; [Ar] 3d6 4s2
Cobalt; [Ar] 3d6 4s2 4p1
Explanation:
We know that in our universe, there are five d orbitals known. However, in this strange universe L, there are only three d orbitals known.
The sixth and seventh first row transition elements are iron and cobalt. In universe L, the electronic configuration of iron and cobalt will be written as;
Iron; [Ar] 3d6 4s2
Cobalt; [Ar] 3d6 4s2 4p1
When an element shares electrons with other atoms of the same or different elements to acquire stable electronic configuration, it is called covalency. If an atom shares 1 electron, its covalency is equal to 1. ... It needs 3 electrons to complete its octet.
Answer:
36.08%
Explanation:
We are given the overall formula of the compound as: CuSO4•5H2O
Now, atomic mass of the elements are;
Cu = 63.55 g/mol
S = 32.07 g/mol
O = 16 g/mol
H20 = 18.02 g/mol
Now, let's calculate the total mass of the compound:
(63.55 g/mol) + (32.07 g/mol) + 4(16 g/mol) + 5(18.02 g /mol) = 249.72 g/mol
From the above, water in the compound is 5(H20)
Thus, total water atomic mass = 5 × 18.02 = 90.1 g/mol
Thus, percentage of water = (atomic mass of water/total mass of compound) × 100%
Percentage of water = (90.1/249.72) × 100% = 36.08%
Answer:
177.277amu
Explanation:
the total occuring isotopes for Hafnium is =6.
First isotope had an atomic weight of 173.940amu
Second isotope =175.941amu
Third isotope =176.943amu
Fourth isotope=177.944amu
Fifth isotope. =178.946amu
sixth isotope .179.947amu
<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>
Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu
Average atomic weight= 1063.661amu /6 = 177.2768333amu
= 177.277amu to 3 decimal places.
Answer:
3.41 x10⁶ torr
Explanation:
To solve this problem we need to remember the equivalency:
1 torr = 133.322 Pa
Then we can proceed to<u> convert 4.55×10⁸ Pa into torr.</u> To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert <em>from</em> in the <em>denominator</em>, and the value we want to convert <em>to</em> in the <em>numerator</em>:
4.55x10⁸ Pa *
3.41 x10⁶ torr