Answer:
[H3O+] = 0.0117 M
[OH-] = 8.5 * 10^-13 M
Explanation:
Step 1: Data given
Concentration of HCl = 0.0117 M
Step 2:
HCl is a strong acid
pH of a strong acid = -log[H+] = - log[H3O+]
[H3O+] = 0.0117 M
pH = -log(0.0117)
pH = 1.93
pOH =14 - 1.93 = 12.07
pOH = -log[OH+] = 12.07
[OH-] = 10^-12.07 = 8.5 * 10^-13
Or
Kw / [H3O+] = [ OH-]
10^-14 / 0.0117 = 8.5*10^-13
I think it might be C because we don't have too much room to dispose of these things with no harm to people
Explanation:
The chemical equation is as follows.

And, the given enthalpy is as follows.
;
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.

x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
Answer:
Pb(No3)2+2Nacl_PbCl2+2NaNO3
50 grams salt
Volume of 50 Grams of Salt
50 Grams of Salt =
2.78 Tablespoons
8.33 Teaspoons
0.17 U.S. Cups
0.14 Imperial Cups