Answer:
Superscript 239 subscript 94 upper P u
(plutonium-239)
Explanation:
In alpha decay, the element subtracts 4 from its mass number and subtracts 2 from its atomic number.
So if we end with Uranium-235, which has an atomic number of 92:
235+4 = 239
92+2 = 94
Element 94 is Pu, plutonium, and its mass could very well be 239 if it's an isotope. So the answer is Superscript 239 subscript 94 upper P u.
Explanation:
The bond between C and O in CO₂ and O and H in H₂O
Therefore,
Option C is correct✔
<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M
1. research question
2. background research
3. hypothesis
4. <span>Controlled experiment
5. data analysis
6. data collection
7. conclusion</span>