HF - The liquid state of this compound will exhibit hydrogen bonding due to the large electronegativity of H and F.
CF4 - The liquid state of this compound will exhibit van der Waals forces.
H2S - The liquid state of this compound will exhibit dipole-dipole or electrostatic attraction forces.
Answer:
![\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}](https://tex.z-dn.net/?f=%5Cmathtt%7B%20t_1-t_2%3D%20In%28%5Cdfrac%7B3%7D%7By%7D%29%20%5Ctimes%20%5Cdfrac%7B7.13%20%5Ctimes%2010%5E8%7D%7BIn2%7D%20%5C%20years%7D)
Explanation:
Given that:
The Half-life of
=
is less than that of ![^{238} U = 4.51 \times 10^9 \ years](https://tex.z-dn.net/?f=%5E%7B238%7D%20U%20%3D%204.51%20%5Ctimes%2010%5E9%20%5C%20years)
Although we are not given any value about the present weight of
.
So, consider the present weight in the percentage of
to be y%
Then, the time elapsed to get the present weight of
= ![t_1](https://tex.z-dn.net/?f=t_1)
Therefore;
![N_1 = N_o e^{-\lambda \ t_1}](https://tex.z-dn.net/?f=N_1%20%3D%20N_o%20e%5E%7B-%5Clambda%20%5C%20t_1%7D)
here;
= Number of radioactive atoms relating to the weight of y of ![^{235}U](https://tex.z-dn.net/?f=%5E%7B235%7DU)
Thus:
![In( \dfrac{N_1}{N_o}) = - \lambda t_1](https://tex.z-dn.net/?f=In%28%20%5Cdfrac%7BN_1%7D%7BN_o%7D%29%20%3D%20-%20%5Clambda%20t_1)
--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of
to be = ![t_2](https://tex.z-dn.net/?f=t_2)
Then:
---- (2)
here;
= Number of radioactive atoms of
relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:
![In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)](https://tex.z-dn.net/?f=In%28%20%5Cdfrac%7BN_o%7D%7BN_1%7D%29%20-In%28%20%5Cdfrac%7BN_o%7D%7BN_2%7D%29%20%3D%20%20%5Clambda%28%20t_1-t_2%29)
replacing the half-life of
=
( since
)
∴
![\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}](https://tex.z-dn.net/?f=%5Cmathtt%7BIn%28%5Cdfrac%7B3%7D%7By%7D%29%20%5Ctimes%20%5Cdfrac%7B7.13%20%5Ctimes%2010%5E8%7D%7BIn2%7D%3D%20t_1-t_2%7D)
The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is ![\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}](https://tex.z-dn.net/?f=%5Cmathtt%7B%20t_1-t_2%3D%20In%28%5Cdfrac%7B3%7D%7By%7D%29%20%5Ctimes%20%5Cdfrac%7B7.13%20%5Ctimes%2010%5E8%7D%7BIn2%7D%20%5C%20years%7D)
Answer:
golfer b had more space giving him the opportunity to gain momentum and strength to strike the ball
You need a list of elements to compare.
The procedure is to wirte the electron configuration and compare the number of electrons in the valence electron shell.
I will work with this list, to show you the procedure
(1) tin
(2) sulfur
(3) arsenic
(4) fluorine
element atomic number number of abbreviated electron
electrons configuration
tin 50 50 [Kr] 5s^2 4d^10 5p2
sulfur 16 16 [Ne] 3s^2 3p^4
arsenic 33 33 [Ar] 4s^2 3d^10 4p^3
fluorine 9 9 [He] 2s^2 2p^5
The valence electrons are those in the last shell.
So, fluorine has 2 + 5 = 7 valence electrons, while tin has 2 + 2 = 4, sulfur has 2 + 4 = 6 and aersenic has 2 + 3 = 5.
Therefore, the answer is fluorine.