Answer:
carbon dioxide and oxygen
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:
7 orbitals
Explanation:
An f sublevel has 7 orbitals
Answer:
C. BF3
Explanation:
The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.
Answer:
The solution is not ideal and shows a positive deviation from Raoult’s law since Psolution (experimental) > Psolution (actual).
Explanation:
Number of moles of CS2 = 3.95g/76.13gmol-1 = 0.0519 moles
Number of moles of acetone = 2.43g/58.08gmol-1 = 0.0418 moles
Total number of moles = 0.0937 moles
Mole fraction of CS2 = 0.0519/0.0937 = 0.5538
Mole fraction of acetone = 0.0418/0.0937 = 0.4461
From Raoult’s law;
PCS2 = 0.5538 × 515 torr = 285.207 torr
Pacetone = 0.4461 × 332 torr = 148.1052 torr
Total pressure = 285.207 torr + 148.1052 torr = 433.3 torr
