(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 
.
(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)
.
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.
Answer: A) mass on earth surface = 5.91kg
B) mass on surface of jupiter = 5.91kg
C) weight on surface of jupiter = 10.697N
Explanation:
The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below
W=mg
From the question, g= 9.8m/s² and weight on the surface on the earth is 58N
A) The mass of watermelon on earth is
m = 58/ 9.8 = 5.91kg
B) the mass of the watermelon on jupiter is 5.91kg.
You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.
C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as
W = mg
W = 5.91 x 9.8
= 10.697N.
Answer:
0.02585 BTU
Explanation:
Given: Spring constant, k = 270 lbf/in
Initial force, f =100 lbf
Compression, x = 1 in
Work done can be calculated as follows:
<span>The element bromine has two isotopes: Br-79 and Br-81, with a 50%-50% isotopic abundance. Statistically, 25% of bromine molecules will be Br79-Br79, 25% will be Br81-Br81 and 50% will be Br79-Br81. This is equivalent to a ratio of 1:1:2 or 1:2:1. The peaks in a mass spectrum just like chromatography reflect this relative abundance of different isotopic combinations.</span>
