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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The organization is based off solitaire and many of the elements are the same
Answer:
\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}
Explanation:
Hello,
The suitable differential equation for this case is:

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

Of course,
.
Now, since the volume of a cone is
and the ratio
or
, the volume becomes:

We proceed to its differentiation:

Then, we compute 

Finally, at h=2:

Best regards.
Answer:
hope it helps you a little
Electrons are the thing in the outer energy levels