1. Physical matter
2. Chemical matter
3. Physical matter
4. Physical matter
5. Chemical matter
6. Physical matter.
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
It is possible to set up the following equilibrium expression:
![K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%3D0.0110)
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of 
(reaction extent) would be:
Which can be solved for to obtain two possible results:
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
![[IBr]=2x=2*0.0249M=0.049M](https://tex.z-dn.net/?f=%5BIBr%5D%3D2x%3D2%2A0.0249M%3D0.049M)
Regards!
Answer:
must have the same number of protons
Answer:
hope this helps :)
Explanation:
for the first one, you can look at the periodic table and look at the atomic number and it will show you how many protons there are giving you the answer because protons and electrons are equal in a pure element
a- carbon
b- neon
c- boron
d- oxygen
e- helium
f- hydrogen
g- lithium
h- beryllium
i- nitrogen
1- sulfur
2- S
3- 16
4- 32.066
5- 16
6- 16
7- 16.066
8- draw circles and put 16 dots like on the other page and in the middle put 16 nuetrons and electrons
9- 6 i think
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
