Answer:
Thus, helium has the largest first ionization energy, while francium has one of the lowest.
Answer:
1. <u>No, you cannot calculate the solubility of X in water at 26ºC.</u>
Explanation:
You cannot calculate the solubility of X in <em>water at 26 degrees Celsius </em>because you do not know whether the solution formed by dissolving the crystals in 3.00 liters of water is saturaed or not.
The only way to determine the solubility of the compound X is by dissolving the crystals in certain (measured) amount of water and making sure that some crystals remain undissolved, as a solid on the bottom of the beaker.
Next, you should filter the solution to remove the undissolved crystals. Then, weigh the solution, evaporate, wash, dry, and weigh the crystals.
Then you have the mass of the crystals dissolved and the mass of the solution which will let you calculate the mass of pure water, and then the solubility.
Electrolysis of water<span> is the </span><span>decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced. Water is separeted into two molecules:
</span>Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻<span> → H</span>₂(g<span>).
</span><span><span>Reaction of oxidation at anode: 2H</span></span>₂<span><span>O(l) → O</span></span>₂<span><span>(g) + 4H</span></span>⁺(<span><span>aq) + 4e</span></span>⁻.<span><span>
</span></span>
Answer:
C.) 2
Explanation:
The pH equation is:
pH = -log[H⁺]
In this equation, [H⁺] is the molarity of the acid. In this case, the acid is HCl. Molarity can be found using the equation:
Molarity (M) = moles / volume (L)
Since you were given moles and volume, you can find the molarity of HCl.
Molarity = moles / volume
Molarity = 0.01 moles / 1.00 L
Molarity = 0.01 M
Now, you can plug the molarity of the acid into the pH equation.
pH = -log[H⁺]
pH = -log[0.01]
pH = 2
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.
