Answer:
3.41 x10⁶ torr
Explanation:
To solve this problem we need to remember the equivalency:
1 torr = 133.322 Pa
Then we can proceed to<u> convert 4.55×10⁸ Pa into torr.</u> To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert <em>from</em> in the <em>denominator</em>, and the value we want to convert <em>to</em> in the <em>numerator</em>:
4.55x10⁸ Pa *
3.41 x10⁶ torr
Answer: 2 (2 neutrons are produced).
Explanation:
1) In the left side of the transmutation equationa appears:
²³⁵U + ¹n →
I am omitting the atomic number (subscript to the leff) because the question does not show them as it is focused on number of neutrons.
2) The right side of the transmutation equation has:
→ ¹⁴⁴Ce + ⁹⁰Sr + ?
3) The total mass number of the left side is 235 + 1 = 236
4) The total mass number of Ce and Sr on the right side is 144 + 90 = 234
5) Then, you are lacking 236 - 234 = 2 unit masses on the right side which are the 2 neutrons that are produced along with the Ce and Sr.
The complete final equation is:
²³⁵U + ¹n → ¹⁴⁴Ce + ⁹⁰Sr + 2 ¹n
Where you have the two neutrons produced.
No. insertions add a nucleotide, deletions delete a nucleotide sequence.
<u>Answer:</u> The equilibrium concentration of
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:

The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:

![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of
is 1.285 M.