The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?
Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
<em>"Your question seems to be missing the correct symbol for the element" </em>
Argon = Ar
Learn more here:brainly.com/question/4628363
Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
Answer:
For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions
For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14
So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions
as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.
Answer:
I think its C but if its not try A then
Explanation:
Sapphire.
It is a stone not a metal.
