The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
Answer:
2 electrons are transfered in this reaction.
Explanation:
Oxidation is a reaction where an atom, ion, or molecule loses electrons, while reduction corresponds to the electron gain of an atom, ion, or molecule.
In an oxidation-reduction reaction two simultaneous processes take place, oxidation and reduction.
So, oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. They are also called electron transfer reactions since the particle that is exchanged is the electron.
In this case:
Zn(s) ⇒ Zn²⁺(aq) + 2 e⁻
2 Ag⁺ (aq) + 2 e⁻ ⇒ 2 Ag(s)
So, zinc metal loses two electrons to form the zinc(II) ions, while the two silver ions each gain one electron to form two silver metal atoms.
Then, Zn is a reducing agent (The reducing agent is the one that provides the electrons, oxidizing itself), AgNO3 is an oxidizing agent (The oxidizing agent is the one that traps the electrons, reducing itself).
Finally, you can see that <u><em>2 electrons are transfered in this reaction.</em></u>
Answer:
Removal of Third Electron
Explanation:
a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.
Ionization energy increases from bottom to top within a group, and increases from left to right within a period.
Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
<em>Where K of equilibrium is the Ka2: 6.5x10⁻⁸</em>
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
<h3>pH = 7.1581</h3>
