Answer:
The hydrosphere.
Explanation:
It is this because ice and glaciers are all liquids and liquids are in the hydrosphere. Well.. Most liquids..
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
Answer:
![[HCl]_{eq}=0.05M](https://tex.z-dn.net/?f=%5BHCl%5D_%7Beq%7D%3D0.05M)
Explanation:
Hello,
In this case, for the given chemical reaction, the law of mass action at equilibrium results:
![Kc=\frac{[H_2]_{eq}[Cl_2]_{eq}}{[HCl]^2_{eq}}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D_%7Beq%7D%5BCl_2%5D_%7Beq%7D%7D%7B%5BHCl%5D%5E2_%7Beq%7D%7D)
Next, in terms of the change
due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:
![4.00=\frac{(x)(x)}{(0.25-2x)^2}](https://tex.z-dn.net/?f=4.00%3D%5Cfrac%7B%28x%29%28x%29%7D%7B%280.25-2x%29%5E2%7D)
Thus, solving for
via quadratic equation or solver, the following results are obtained:
![x_1=0.1M\\x_2=0.17M](https://tex.z-dn.net/?f=x_1%3D0.1M%5C%5Cx_2%3D0.17M)
Clearly, the solution is
as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:
![[HCl]_{eq}=0.25M-2*0.1M](https://tex.z-dn.net/?f=%5BHCl%5D_%7Beq%7D%3D0.25M-2%2A0.1M)
![[HCl]_{eq}=0.05M](https://tex.z-dn.net/?f=%5BHCl%5D_%7Beq%7D%3D0.05M)
Best regards.
Answer:
The answer to your question is : 16.9 g of Ag
Explanation:
Data
26 g Ag
10.8 g of Sn
2.4g Cu
0.8 Zn
Ag = ? in 26 g of sample
Total mass in the amalgam = 26 + 10.8 + 2.4 + 0,8 = 40 g
Rule of three
40 g of sample -------------- 26 g of silver
26 g of sample -------------- x
x = (26 x 26) / 40
x = 16.9 g of Silver