Answer:
pH = 1.95
Explanation:
For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.
Therefore... the [H⁺] concentraion and pH are computed as follows...
[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M
pH = -log[H⁺] = -log(0.01118) = 1.95
Answer:
TATGGCGTT
Explanation:
Complimentary base pairs:
A-T
C-G
Use the other letter for complimentary strands
Making repeated separations of the various substances in the pitchblende, Marie and Pierre used the Curie electrometer to identify the most radioactive fractions. They thus discovered that two fractions, one containing mostly bismuth and the other containing mostly barium, were strongly radioactive.
<h3>What was surprising about pitchblende?</h3>
Since it was no longer appropriate to call them “uranic rays,” Marie proposed a new name: “radioactivity.”
Even more surprising, Marie next found that a uranium ore called pitchblende contained two powerfully radioactive new elements: polonium, which she named for her native Poland, and radium.
<h3>Why is radium more radioactive than uranium?</h3>
It is 2.7 million times more radioactive than the same molar amount of natural uranium (mostly uranium-238), due to its proportionally shorter half-life.
Learn more about highly radioactive elements here:
<h3>
brainly.com/question/10257016</h3><h3 /><h3>#SPJ4</h3>
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
