K because parent atoms are always larger than their cations(positively charged atoms)
<u>Question:</u>
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K
<u>Answer:</u>
The mean activity coefficient for HCl solution is 0.78.
<u>Explanation:</u>
Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is
As we know that = 0.22 V and E = 0.342 V, the equation will become
So, the mean activity coefficient is 0.78.
<span>2 Na + CaF</span>₂<span> = 2 NaF + Ca
is an example of a </span>single- replacement
answer D
hope this helps!
Answer:
Lead Pb
Explanation:
Firstly, we need to know what occurs when a radioisotope emits an alpha particle. An alpha particle is an helium atom. When an isotope emits an alpha particle, it loses an helium atom corresponding to subtracting 4 from its mass number and 2 from its atomic number. This of course coupled with the release of radiation.
Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.
Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.
210Po ——> 206Pb + alpha particle + radiation
Answer:
At equilibrium, the concentration of is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+ ⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor present. Immediately, and are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: []=0 ; [ ]= 0 ; []=0.60M
C: []=+x ; [ ]= +x ; []=-2x
E: []=0+x ; [ ]= 0+x ; []=0.60-2x
Now we can use the constant information:
=
=
=
At equilibrium, the concentration of is going to be 0.30M