Answer:
cation +2
Explanation:
A cation is a positively charged ion, while an anion is a negatively charged ion. Contrary to how it sounds, when an atom loses electrons, it's charge becomes more positive. Since the calcium atom lost 2 electrons, it is a positive (+2) ion.
D. Aquifers
Hope this helps chief.
<u>Answer:</u> The molar mass of the gas is 35.87 g/mol.
<u>Explanation:</u>
To calculate the mass of gas, we use the equation given by ideal gas:
PV = nRT
or,

where,
P = Pressure of gas = 945 mmHg
V = Volume of the gas = 0.35 L
m = Mass of gas = 0.527 g
M = Molar mass of gas = ? g/mo
R = Gas constant = 
T = Temperature of gas = ![88^oC=[88+273]=361K](https://tex.z-dn.net/?f=88%5EoC%3D%5B88%2B273%5D%3D361K)
Putting values in above equation, we get:

Hence, the molar mass of the gas is 35.87 g/mol.
The cell notation is
║
and the cell potential is 0.464
The reaction occurred while losing of hydron is known as oxidation reaction
We can also tell that the reaction occurred while gaining of oxygen atom is known as oxidation reaction.
The reaction occurred while gaining of hydrogen is known as reduction reaction or we can say that the reaction occurred while losing oxygen atom is known as reduction reaction
An electrochemical cell's cell potential is defined as the difference in potential between two half cells. The electrons' capacity to go from one half cell to the other is what causes the potential difference. As a result of the chemical reaction being a redox reaction, electrons can travel across electrodes.
Calculating the Cell potential
E°cell = E°(reduction) - E°(oxidation)
= 0.34 - (-0.0124)
= 0.464
Hence the cell potential is 0.464
Learn more about Cell potential here
brainly.com/question/11638563
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Answer:
pH = 2.21
Explanation:
Hello there!
In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:
![[HF]=\frac{0.450mol}{1.0L} =0.450M](https://tex.z-dn.net/?f=%5BHF%5D%3D%5Cfrac%7B0.450mol%7D%7B1.0L%7D%20%3D0.450M)
![[NaF]=\frac{0.050mol}{1.0L} =0.050M](https://tex.z-dn.net/?f=%5BNaF%5D%3D%5Cfrac%7B0.050mol%7D%7B1.0L%7D%20%3D0.050M)
Now, calculated the pKa of HF:

We can proceed to the HH equation:
![pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BNaF%5D%7D%7B%5BHF%5D%7D%20%29%5C%5C%5C%5CpH%3D3.17%2Blog%28%5Cfrac%7B0.05M%7D%7B0.45M%7D%20%29%5C%5C%5C%5CpH%3D2.21)
Best regards!