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3 years ago

What happens when magnesium reacts with iodine to make magnesium iodide?

Chemistry

1 answer:

I am Lyosha [343]3 years ago

7 0

Explanation:

Atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. On the other hand, atomic number of iodine is 53 and its electronic configuration is $[Kr]4d^{10}5s^{2}5p^{5}$ .

Hence, there are 7 valence electrons in an iodine atom and there are 2 valence electrons in a magnesium atom.

So, one atom of iodine requires one electron from a donor atom to complete its octet. But one magnesium atom contains two valence electrons.

Therefore, one magnesium atom will combine with two iodine atoms to result in the formation of magnesium iodide as follows.

$Mg^{2+} + 2I^{-} \rightarrow MgI_{2}$

Therefore, an ionic bond will be formed when magnesium reacts with iodine to make magnesium iodide.

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n the periodic table, the atomic mass of oxygen is listed as 15.9994 amu. Based on this, one can deduce that the most common ___

Anit [1.1K]

Isotope- variation of an element

Sixteen- atomic number of oxygen

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3 years ago

Read 2 more answers

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

Answer: The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

Explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

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3 years ago

Why can't the reaction, ZnCl2 + H2 → Zn + 2HCI, occur naturally?

Llana [10]

Answer:

It is fairly obvious that zinc metal reacts with aqueous hydrochloric acid! The bubbles are hydrogen gas. ... In fact, electrons are being transferred from the zinc atoms to the hydrogen atoms (which ultimately make a molecule of diatomic hydrogen), changing the charges on both elements.

Explanation:

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3 years ago

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

The system is not in equilibrium and will evolve left to right to reach equilibrium.

3 0

3 years ago

The dehydration of the alcohol functional group is a widely used reaction in organic chemistry. The mechanism is generally accep

puteri [66]

Answer:

Elimination
Elimination
Zaitsev
Zaitsev
Carbocation

Explanation:

The mechanism is generally accepted to always operate via an ELIMINATION step-wise process.

The ELIMINATION mechanism process will always produce (after dehydration) a ZAITSEV style alkene as major product

The driving force for the production of this ZAITSEV style alkene product is generally going to be determined by stability of the CARBOCATION

Elimination mechanism is the removal of two substituents from a molecule in either a one- or two-step mechanism

Carbocation is a molecule containing a positive charged carbon atom and three bonds

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3 years ago