<span>First, write the net ionic equation for the unbalanced reaction. If you are given a word equation to balance, you'll need to be able to identify strong electrolytes, weak electrolytes and insoluble compounds. Strong electrolytes completely dissociate into their ions in water. Examples of strong electrolytes are strong acids, strong bases, and soluble salts. Weak electrolytes yield very few ions in solution, so they are represented by their molecular formula (not written as ions). Water, weak acids, and weak bases are examples of weak electrolytes. The pH of a solution can cause them to dissociate, but in those situations, you'll be presented an ionic equation, not a word problem. Insoluble compounds do not dissociate into ions, so they are represented by the molecular formula. A table is provided to help you determine whether or not a chemical is soluble, but it's a good idea to memorize the solubility rules.
</span><span><span>arate the net ionic equation into the two half-reactions. This means identifying and separating the reaction into an oxidation half-reaction and a reduction half-reaction. </span><span>For one of the half-reactions, balance the atoms except for O and H. You want the same number of atoms of each element on each side of the equation. </span><span>Repeat this with the other half-reaction. </span><span>Add H2O to balance the O atoms. Add H+ to balance the H atoms. The atoms (mass) should balance out now. </span><span>Now balance charge. Add e- (electrons) to one side of each half-reaction to balance charge. You may need to multiply the electrons the the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation. </span><span>Now, add the two half-reactions together. Inspect the final equation to make sure it is balanced. Electrons on both sides of the ionic equation must cancel out. </span><span>Double-check your work! Make sure there are equal numbers of each type of atom on both sides of the equation. Make sure the overall charge is the same on both sides of the ionic equation. </span><span>If the reaction takes place in a basic solution, add an equal number of OH- as you have H+ ions. Do this for both sides of the equation and combine H+ and OH- ions to form H2O. </span><span>Be sure to indicate the state of each species. Indicate solid with (s), liquid for (l), gas with (g), and aqueous solution with (aq). </span><span>Remember, a balanced net ionic equation only describes chemical species that participate in the reaction. Drop additional substances from the equation.ExampleThe net ionic equation for the reaction you get mixing 1 M HCl and 1 M NaOH is:H+(aq) + OH-(aq) → H2O(l)Even though sodium and chlorine exist in the reaction, the Cl- and Na+ ions are not written in the net ionic equation because they don't participate in the reaction.</span></span>
Answer:
It evaporates and moves into the air.
Explanation:
When water is left out for a while, it evaporates into the air! :)
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
Answer:HNO₃ and NO³⁻ would not function as buffer
Explanation:
The buffer solution are usually prepared by using any weak acid (which would partially dissociate) and mixing this weak acid with its own conjugate base or any weak base (which would partially dissociate) and mixing with with its conjugate acid.
A buffer solution is a solution which resists change in pH of the solution.
Since nitric acid is a very strong acid and hence neither nitric acid HNO₃ or its conjugate base NO³⁻ anionb is suitable for the preparation of buffer solution.
HCO³⁻ is a weak acid and hence it can form a buffer solution with its conjugate base CO₃²-. so they can be used to form buffer.
C₂H₅COOH is a weak acid and hence it can also form buffer solution with its conjugate base.
So only HNO₃and NO³⁻ would not be able to form buffer
So option a is the answer.
Gold has 79 protons on the periodic table
