A student is performing experiments on a particular substance. Which

For each element, predict where the "jump " occurs for successive ionization energies. (For example, does the jump occur between

vichka [17]

Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron

The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

The electronic configuration of Lithium is 1s2 2s1

There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.

8 0

4 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

In which set of elements would all members be expected to have very similar chemical properties?

zepelin [54]

Answer:

each group of the periodic table.

example group 7 - fluorine, chlorine, bromine, iodine, astatine, tennessine

6 0

2 years ago

2zn+o2=2zno how many moles of zinc are needed to make 6 moles of zinc oxide?

Temka [501]

Answer:

$\boxed{\text{6 mol}}$

Explanation:

(a) Balanced equation

2Zn + O₂ ⟶ 2ZnO

(b). Calculation

You want to convert moles of ZnO to moles of Zn

The molar ratio is 2 mol Zn:2 mol ZnO

$\text{Moles of Zn} =\text{6 mol ZnO} \times \dfrac{\text{2 mol Zn}}{\text{2 mol ZnO}} = \text{6 mol Zn}\\\\\text{You need }\boxed{\textbf{6 mol of Zn}}\text{ to form 6 mol of ZnO}.$

3 0

3 years ago

Balance each of the following redox reactions occurring in acidic aqueous solution.

dmitriy555 [2]

Answer:

Part A : Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

Part B : 3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).

Part C: 3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.

Explanation:

Part A : Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s), Express your answer as a chemical equation. Identify all of the phases in your answer.

It is balanced as written: Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

The two half reactions are:

The oxidation reaction: Zn(s) → Zn²⁺(aq) + 2e.

The reduction reaction: Sn²⁺(aq) + 2e → Sn(s).

To obtain the net redox reaction, we add the two-half reactions as the no. of electrons in the two-half reactions is equal.

So, the net chemical equation is:

Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

Part B: Mg(s) + Cr³⁺(aq) → Mg²⁺(aq) + Cr(s), Express your answer as a chemical equation. Identify all of the phases in your answer.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: Mg(s) → Mg²⁺(aq) + 2e.

The reduction reaction: Cr³⁺(aq) + 3e → Cr(s).

To obtain the net redox reaction, we multiply the oxidation reaction by 3 (3Mg(s) → 3Mg²⁺(aq) + 6e) and the reduction reaction by 2 (2Cr³⁺(aq) + 6e → 2Cr(s)) to equalize the no. of electrons in the two-half reactions.

So, the net redox reaction will be:

3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).

Part C : MnO⁴⁻(aq) + Al(s) → Mn²⁺(aq) + Al³⁺(aq), Express your answer as a chemical equation. Identify all of the phases in your answer.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: Al → Al³⁺ + 3e.

The reduction reaction: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O.

To obtain the net redox reaction, we multiply the oxidation reaction by 5 (5Al → 5Al³⁺ + 15e) and the reduction reaction by 3 (3MnO₄⁻ + 24H⁺ + 15e → 3Mn²⁺ + 12H₂O) to equalize the no. of electrons in the two-half reactions.

So, the net redox reaction will be:

3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.

4 0

3 years ago