Answer:
16.8%
Explanation:
31% NaOH molar mass 40 gm
69% H2O molar mass 18 gm
1000 gm would be
310 gm NaOH or 310/40 = 7.75 moles
690 gm of H2O or 690/18 = 38.333 moles
7.75 / (7.75 + 38.333) = .168 mole fraction
Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = ![\frac{[H+][IO-]}{[HIO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%2B%5D%5BIO-%5D%7D%7B%5BHIO%5D%7D)
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density
The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.
r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.
r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.
d(NaF) = r(Na⁺) + r(F⁻).
d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.
d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.
The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
