Answer:
Yes it would
Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition
Answer:
John Dalton
Explanation:
Both John Dalton and Democritus thought that the atom was an indivisible sphere until J.J. Thompson came out with the plum pudding model. Hope I helped!
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.
Best regards!
K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent