Answer:
I believe it is B,fluorine to complete the octet rule
Explanation:
The answer would be D because from my research it's the only one that didn't have a catalyst
Answer:
The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.
Explanation:
Number of hours worker exposed to xylene = 
The concentration of xylene in the workplace =
The worker is inhaling air at a rate of
.
Amount xylene inhaled by worker in an hour :
= 
Amount xylene inhaled by worker in 320 hours:

1 μg = 0.001 mg
Amount xylene inhaled by worker in 320 hours = 11.520 mg
1 day = 24 hours
Amount xylene inhaled by worker in 1 day:

Assuming 70 kg body mass, the chronic daily intake of xylene :

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.
Answer:
2K +F₂→ 2KF
Explanation:
When we balance an equation, we are trying to ensure that the number of atoms of each element is the same on both sides of the arrow.
On the left side of the arrow, there is 1 K atom and 2 F atoms. On the right, there is 1 K and 1 F atom.
Since the number of K atoms is currently balanced, balance the number of F atoms.
K +F₂→ 2KF
Now, that the number of F atoms is balanced on both sides, check if the number of K atoms are balanced.
<u>Left</u>
K atoms: 1
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The number of K atoms is not balanced.
2K +F₂→ 2KF
<u>Left</u>
K atoms: 2
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The equation is now balanced.
Explanation:
These elements are rare because:
<u>Helium fuses into the carbon by the combination of three helium nuclei (Z = 2) and one carbon nucleus (Z = 6), therefore bypassing elements with Z= 3, 4 and 5 which are lithium, beryllium, and boron respectively. Therefore, the fusion processes in cores of the stars do not form these three elements. </u>