The concentration of hydrogen ions in a solution is a measure of its acidity. So the correct option is (b) false.
When an Arrhenius acid is dissolved in water, hydrogen ions are produced:
H+(aq) + A- = HA + H2O (aq)
Here, H+ is the hydrogen cation, A- is the solvated anion, also known as the conjugate base, and HA is the non-dissociated acid. When an Arrhenius base is dissolved in water, hydroxide ions are produced:
BOH + H2O → B+(aq) + OH-(aq)
Is a material with at least one hydrogen atom that has the ability to split apart in an aqueous solution to produce an anion and an H + ion (a proton), creating an acidic solution. Bases are substances that, when dissolved in water, create hydroxide ions (OH) and a cation, resulting in a basic solution.
Learn more about hydrogen here:
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Answer:
I think option A is right answer
D. Potential is not moving, while kinetic is moving.
Answer:
<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of <u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>o</u><u>l</u><u>:</u>
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<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.
Answer:
The answer to your question is 6.0 moles of O₂
Explanation:
Data
2KClO₃ ⇒ 2KCl + 3O₂
moles of O₂ = ?
moles of KCl = 4
Process
To find the number of moles of O₂, use proportions and cross multiplication.
Use the coefficients of the balanced equation.
2 moles of KCl ----------------- 3 moles of O₂
4 moles of KCl ----------------- x
x = (4 x 3) / 2
-Simplification
x = 12/2
-Result
x = 6 moles of O₂
-Conclusion
When 4,0 moles of KCl are produced, 6.0 moles of O₂ will be produced.