Answer:
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Density decreases that's why ice floats on water because it's less dense than water.
Answer:
- <u><em>Option D. has a great [OH⁻]</em></u>
Explanation:
1) Both <em>acids</em> and <em>bases</em> ionize in aqueous solutions so they are able to <em>conduct electricity</em>.
The ions, being charged particles, when flow through the solution are charge carriers, then they conduct electricity.
So, the option A does not state a difference between a solution of a base and a solution of an acid.
2) Both acids and bases are able to cause an <em>indicator color change</em>.
The usufulness of the indicators is that they are able to change of color when the pH changes either from acid to basic or from basic to acid. There are different indicators because none is suitable for the whole range of pH, but the statement B is not how solutions of base and acids differ.
3) The model of Arrhenius for acids and bases states that an acid is a substance that ionizes in water releasing H⁺ ions (this is equivalent to H₃O⁺) and a base is a substance that releases OH⁻ ions in water. Then, acids have a greater concentration of H₃O⁺ (so option C is not true for a solution of a base) and bases have a greater concentraion of OH⁻, making the option D. true.
To combine these you would need to perform Dehydration synthesis which removes a water molecule. So it would be c12h22o11
Answer:
- Mass of monobasic sodium phosphate = 1.857 g
- Mass of dibasic sodium phosphate = 1.352 g
Explanation:
<u>The equilibrium that takes place is:</u>
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:
pH = pka + ![log\frac{[A^{-} ]}{[HA]}](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:
![7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]](https://tex.z-dn.net/?f=7.0%3D7.21%2Blog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%20-0.21%3Dlog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%5C%5C10%5E%7B-0.21%7D%20%3D%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C0.616%20%2A%20%5BH2PO4%5E%7B-%7D%5D%20%3D%20%5BHPO4%5E%7B-2%7D%5D)
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
- Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
- Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.
- mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
- mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g