Answer:
Explanation:
The acidity of a solution is measured by its pH, which is the logarithm of the inverse of the molar concentration of hydronium (H₃O⁺) ions:
- pH = log 1 / [H₃O⁺] = - log [H₃O⁺]
When you know the pH value you can find hydronium concentration using the antilogaritm function:
![pH=-log[H_3O^{+}]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-2.50}\\ \\ {[H_3O^+]}=0.0032](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-pH%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-2.50%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D0.0032)
The unit of molar concentration is M.
To prove your answer you can take the logarithm of 0.0316:
Answer:
To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy. Let me know if this helped?
Answer: they are both at the same concentration
Explanation: You will know that the amount of solvent in and around the cell will be equivalent when they have the same amount of concentration. The answer to the question is they are both at the same concentration.
Answer:
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Explanation:
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:

Ionic form of the above equation follows:

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:

Hence, the net ionic equation is written above.