Protons, neutrons, and electrons.
Answer: The pressure after the tire is heated to 17.3°C is 167 kPa
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:

Putting values in above equation, we get:

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa
Answer:

Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:

So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
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When acids react with bases they produce salt and water such as:
HCl + NaOH → NaCl + H₂O
According to strength of acid and base, we have 4 types of salts:
salt of strong acid and strong base like: NaCl
salt of weak acid and strong base like: CH₃COONa
salt of strong acid and weak base like: NH₄Cl
salt of weak acid and weak base like: CH₃COONH₄
Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.