Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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Answer:
Explanation:
6CO₂ + 6 H₂O ⇄ C₆H₁₂0₆ + 6O₂
This is the chemical equation given .
1. The equation shows a __Chemical equation_______the breaking and forming of chemical bonds that leads to a change in the composition of matter.
2. In the equation, CO₂ is a___reactant_____.
3. In the equation, C₆H₁₂0₆ is a ___product________.
4. In O₂, the type of bond that holds the two oxygen atoms together is a_nonpolar_covalent bond_________.
5. In H₂O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a__polar_hydrogen bond____.
6. The number of oxygen atoms on the left side of the equation is__equal to_________ the number of oxygen atoms on the right side.
Answer:
Explanation:
MW of NaOH = 40 g/mol
MW of KCl = 74.55 g/mp;
250 mL = .25 L
100g NaOH / 40 g = 25 mol
100g KCl/ 74.55g = 1.34 mol
Molarity of NaOH: 25/.25 = 100M
Molarity of KCl: 1.34/.25 = 5.36 M
francium , in the Periodic table the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. making helium is the smallest element, and francium the largest.
