Answer:
677
Explanation:
took the test
it said that it was wrong but then it also said the total bond energy of the reactants is 677 kJ/mol. the correct one was 854, i guess.
Which unit is not a measurement of volume?
<u>cm
</u>
L
ml
cm³
Cm is a measurement for length
Answer: E. It decreases the value of Eact.
Explanation:
Activation energy is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.
A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more molecules can cross the energy barrier and more molecules convert to products.
The catalyst itself does not take part in the chemical reaction and is regenerated as such at the end.
Answer:
The concentrations of hydronium and hydroxide ions in a solution with pH 2.6 are 2.51*10⁻³ and 3.98*10⁻¹² respectively.
Explanation:
pH is a measure of the acidity or alkalinity of an aqueous solution. The pH indicates the concentration of hydrogen ions present in certain solutions. Mathematically it is defined as the opposite of the base 10 logarithm or the negative logarithm of the activity of hydrogen ions, whose equation is
pH= -log [H⁺]
So, being pH= 2.6 and replacing in the definition of pH:
2.6= -log [H⁺]
![[H^{+} ]=10^{-2.6}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%3D10%5E%7B-2.6%7D)
[H⁺]=2.51*10⁻³
pOH is a measure of the concentration of hydroxide ions (OH⁻). The sum between the pOH and the pH results in 14:
pOH + pH= 14
So, being pH= 2.6:
pOH + 2.6= 14
pOH= 14 - 2.6
pOH= 11.4
The pOH is defined as the negative logarithm of the activity of the hydroxide ions. That is, the concentration of OH⁻ ions:
pOH= - log [OH⁻]
Being pOH= 11.4 and replacing
11.4= -log [OH⁻]
![[OH^{-} ]=10^{-11.4}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D10%5E%7B-11.4%7D)
[OH⁻]= 3.98*10⁻¹²
<u><em>The concentrations of hydronium and hydroxide ions in a solution with pH 2.6 are 2.51*10⁻³ and 3.98*10⁻¹² respectively.</em></u>