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2 years ago

What is the expected oxidation state for the most common ion of element 2

Chemistry

1 answer:

lozanna [386]2 years ago

3 0

Answer: 1+

Justification:

The ionization energies tell the amount of energy needed to release an electron and form a ion. The first ionization energy if to loose one electron and form the ion with oxidation state 1+, the second ionization energy is the energy to loose a second electron and form the ion with oxidation state 2+, the third ionization energy is the energy to loose a third electron and form the ion with oxidation state 3+.

The low first ionization energy of element 2 shows it will lose an electron relatively easily to form the ion with oxidations state 1+.

The relatively high second ionization energy (and third too) shows that it is very difficult for this atom to loose a second electron, so it will not form an ions with oxidation state 2+. Furthermore, given the relatively high second and third ionization energies, you should think that the oxidation states 2+ and 3+ for element 2 never occurs.

Therefore, the expected oxidation state for the most common ion of element 2 is 1+.

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One way to determine the degree of saturation of a solid-liquid solution is to drop a crystal of the solute into the solution.

Dafna1 [17]

The answer is D: Saturated.

A saturated solution is one in which the exact maximum amount of solute has been dissolved. So, new solute will not dissolve in the solution. In contrast, an unsaturated solution can hold more solute, so if that option were correct, the crystal would have dissolved.

The other two terms are a bit more complicated. A supersaturated solution is one holding an amount of solute above the sustainable limit. Because of that, when more solute is added, the solution will immediately adjust, and some solute will come out of solution in a precipitate. Because the crystal isn't growing, we can eliminate this option.

A concentrated solution is one holding a relatively large amount of solute. However, you can have concentrated solutions that are saturated and unconcentrated (the word for this is dilute) solutions that aren't saturated. Therefore, we can say that because the crystal doesn't dissolve, this solution is saturated, but we can't say with certainty that it is concentrated.

Because the first three options are invalid, as described above, while the scenario does describe a saturated solution, D is the correct answer.

7 0

3 years ago

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

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2 years ago

Choose the correct coefficients for the equation? ____NaCI>____Na+___C12

Wewaii [24]

Explanation:

Ca + NaCl ----> CaCl2 + Na

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2 years ago

At equilibrium the reactant and product concentrations are constant because a change in one direction is balanced by a change in

Anna [14]

Answer: a. The concentrations of the reactants and products have reached constant values

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions. For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equal to rate of the backward reaction.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time and are constant.

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For a equilibrium reaction,

$A\rightleftharpoons B$

$K_{eq}=\frac{[B]}{[A]}$

Thus the correct answer is the concentrations of the reactants and products have reached constant values.

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2 years ago

A compound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?

RUDIKE [14]

The answer is PI3
Ratio of 1:3

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