(a)
Electronic configuration is given as follows:
![[Kr]4d^{3}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B3%7D)
Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.
Thus, electronic configuration of neutral atom is ![[Kr]4d^{5}5s^{1}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B5%7D5s%5E%7B1%7D)
.
The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.
This corresponds to element: molybdenum. Thus, the tripositive atom will be 
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(b) The given electronic configuration is ![[Kr]5s^{2}4d^{2}](https://tex.z-dn.net/?f=%5BKr%5D5s%5E%7B2%7D4d%5E%7B2%7D)
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The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.
This corresponds to element zirconium, represented by symbol Zr.
The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.
The sun has orbited along time so when they ask theses questions I give you the right answer I think lol
<span>A. frustration-aggression theory.
Good Luck!!</span>
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
