M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.
Iron doesn't fit because it doesn't have enough atoms or protons in its nucleus so there for it belongs in column 2. <span />
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25 8.1
25/2 =12.5 + 8.1
12.5/2= 6.25 +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8 of its original mass.
(good luck on the regent if thats what your studying for :)
Answer:

Explanation:
Hello,
In this case, given the reaction:

The total consumed gallons are computed by considering 686 miles were driven and the consumption is 21.2 miles per gallon, thus:

Hence, with the given density, one could compute the consumed grams and consequently moles of gasoline as well as moles that were consumed:

Next, since gasoline (molar mass = 114 is in a 2:16 molar relationship with the yielded carbon dioxide, we compute its produced moles as shown below:

Finally, we could assume the given STP conditions to compute the volume of carbon dioxide, as no more information regarding the space wherein the carbon dioxide is available:

Best regards.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
![ka = \frac{[H_3O^+][X^-]}{[HX]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BX%5E-%5D%7D%7B%5BHX%5D%7D)
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
![[H_3O^+]} = 10^{-2,40}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%7D%20%3D%2010%5E%7B-2%2C40%7D)
[H₃O⁺] = <em>0,004M</em>
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = <em>0,085M</em>
![ka = \frac{[0,004][0,004]}{[0,085]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5B0%2C004%5D%5B0%2C004%5D%7D%7B%5B0%2C085%5D%7D)
ka = 1,88x10⁻⁴
And <em>pka = 3,73</em>
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
<em>pkb = 10,27</em>
I hope it helps!