Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
Answer: Alpha radiation
Explanation: Alpha decay : When a larger radioactive nuclei decays into smaller nuclei by releasing alpha radiation, the mass number and atomic number is reduced by 4 and 2 units respectively.

Beta decay : When a larger radioactive nuclei decays into smaller nuclei by releasing beta radiation, the atomic number is increased by 1 unit.

Gamma decay : When a larger radioactive nuclei decays into smaller nuclei by releasing gamma radiation, the mass number remains same.

<span><em><u>Climatology </u>is a <u>subspecialty </u>of a </em><u><em>Climate</em></u><em> and for how the <u>climate</u> changes. This is averaged out from over a set of a period of time.
<u>I hope this helps! ;D</u></em></span>
Answer:
A Li2O
Explanation:
Li has a 1+ charge and O has a 2- charge so to balance the charges there needs to be 2 Li for every 1 O
Answer:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Explanation:
Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.
3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)
The complete ionic equation includes all the ions and insoluble species.
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)