Answer:
Explanation:
NaCl does not contain molecules
Answer:
Br
|
Br-P-Br
|
Br
Explanation:
To calculate the valance electrons, look at the periodic table to find the valance electrons for each atom and add them together. P is in column 5A, so it has 5, Br is in column 7A, so it has 7 (multiply by 4 since there are 4 Br atoms to give 28) and there is a 1- charge, so add one more electron. 5+28+1=34, so there are 34 electrons to place. P would be the central atom, so place it in the middle. Place each Br around the P (as shown above) with a a single line connecting it. Each line represents 2 electrons, so 8 total have been place, leaving 26 remaining. Place 6 electrons around each Br (2 on each of the unbonded sides), which leaves 2 electrons remaining. The remaining pair of unbound electrons will be attached to the P between any two Br atoms. Phosphorus doesn't have to follow the octet rule, so it actually ends up with 10 valance electrons.
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 = 
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Answer:
Sodium Fluoride F 18 Injection is a positron emitting radiopharmaceutical, no-carrier added.
Explanation:
Element Name Fluorine
Element Symbol F
Atomic Number 9
