Answer:
Polonium: <em>Number</em><em> </em><em>of</em><em> </em><em>protons</em><em> </em><em>=</em><em> </em><em>8</em><em>2</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Number</em><em> </em><em>of</em><em> </em><em>neutrons</em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em>6</em>
Sodium: <em>Number</em><em> </em><em>of</em><em> </em><em>protons</em><em> </em><em>=</em><em> </em><em>1</em><em>1</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Number</em><em> </em><em>of</em><em> </em><em>neut</em><em>rons</em><em> </em><em>=</em><em> </em><em>2</em><em>2</em>
Answer:
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Explanation:
Let's consider the following oxidation half-reaction that takes place in basic aqueous solution.
Mn(s) ⇒ MnO₄⁻(aq)
First, we will perform the mass balance. We will add 4 H₂O to the products side and 8 OH⁻ to the reactants side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l)
Finally, we will perform the charge balance by adding 7 electrons to the products side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Answer:
Yes.
Explanation:
Yes, the change in cuvette lead to a determinate error in the analysis if the different cuvette is used for the analysis because the amount of liquid sample that is used has different volume. If both cuvette are of the same type and has no difference in their structure and size then there is error occurs in the analysis but if both cuvette are different from one another then the error will occur in the analysis. because the amount of liquid that is used has different volume.
For a solution to be tested in this experiment and must be buffered to pH of 10, the [OH-] change will be, The hydroxide concentration would be higher and the formation of insoluble hydroxide salts with Mg^2 and Ca^2 would cause the determined concentration of water hardness to be too low.
<h3>What would be the [OH-] change?</h3>
Generally, the equation for the Total hardness is mathematically given as

Where

D=0.02/0.01=2
Therefore

T=1632ppm
in conclusion, The hydroxide concentration would be higher .
Read more about Concentration
brainly.com/question/16979235