Answer: Inversely , Directly
Explanation:
The energy of a photon is inversely proportional to its wavelength and directly proportional to its frequency.
As can be seen from this equation;
E = hv = h c / ∧
Where E = Energy of a photon
v = Frequency
h = Planck Constant
c = speed of light
∧ = Wave length
Answer:
3 NH4OH (l) + H3PO4 (aq) → (NH4)3PO4 (aq) + 3 H2O (l)
Explanation:
This is an acid-base reaction (neutralization): NH4OH is a base, H3PO4 is an acid
4I+5O2=I4O10? which would be a synthesis reaction
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%
Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!