<u>Given information:</u>
Mass of NaCl (m) = 87.75 g
Volume of solution (V) = 500 ml = 0.5 L
Molar mass of NaCl (M) = 58.44 g/mol
<u>To determine:</u>
The molarity of NaCl solution
<u>Explanation:</u>
Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)
i.e. M = moles of solute/liters of solution = n/V
Moles of solute (n) = mass of solute (m)/molar mass (M)
moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles
Therefore,
Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M
<u>Ans: (D)</u>
The correct answer is option 3. The IUPAC name is Iron(II) sulfide. It is the less stable amorphous form. When this is powdered, it is pyrophoric or it ignites spontaneously in air. It readily reacts with hydrochloric acid producing hydrogen sulfide.
Answer:
boring
Explanation:
life would be meaningless like everyone being like on person
To answer this question, we should give the standard free energy change of ATP to ADP conversion that is equal to 30.5 kJ/mol ATP. In this case, 2400 kcal is equal to 10041. 6 kJ. Hence, the mol of ATP produced is 329.23 mol ATP or equal to 166.93 kg ATP.
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!