What has a complete ionic equation of Mg(s) + Zn2+ + 2NO3 → Mg2+ + 2NO3 + Zn(s)?
O A. Mg(s) + Zn2+ → Mg2+ + Zn(s)
O B. Mg(s) + Zn(NO3)2(aq) + Mg(NO3)2(aq) + Zn(s)
O C. Mg(NO3)2 + Zn(s) → Zn2+ + 2NO3 + Mg(s) O D. Mg(s) + 2Zn(NO3)(aq) + 2MgNO3(aq) + Zn(s)​

Answers

Answer 1

Answer: [tex]Mg(s)+Zn(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Zn(s)[/tex]

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we do not not include the spectator ions in the equations.

Spectator ions are the ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The equation given is:

[tex]Mg(s)+Zn^{2+}+2NO_3^-\rightarrow Mg^{2+}+2NO_3^-+Zn(s)[/tex]

The complete molecular equation will be:

[tex]Mg(s)+Zn(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Zn(s)[/tex]  

 

Answer 2

The complete ionic equation of[tex]\rm Mg(s) + Zn^{2+} + 2NO_3 \rightarrow Mg^{2+} + 2NO_3 + Zn(s)[/tex] is[tex]\rm Mg(s) + Zn^{2+} \rightarrow Mg^{2+} + Zn(s)[/tex]. So, option A is right.

A complete ionic equation shows all of the ions that are present in a reaction, including the spectator ions. In this reaction, magnesium metal (Mg(s)) reacts with zinc(II) ions ([tex]\rm Zn^2+[/tex]) to form magnesium(II) ions ([tex]\rm Mg2+[/tex]) and zinc metal (Zn(s)).

The balanced molecular equation for this reaction is:

[tex]\rm Mg(s) + Zn(NO_3)_2(aq) \rightarrow Mg(NO_3)_2(aq) + Zn(s)[/tex]

The complete ionic equation is:

[tex]\rm Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)[/tex]

The spectator ions are the nitrate ions ([tex]\rm NO_3^-[/tex]), which are present on both sides of the equation. They do not participate in the reaction, so they are not included in the complete ionic equation.

The other choices are incorrect because they do not show all of the ions that are present in the reaction.

Therefore, the answer is A. [tex]\rm Mg(s) + Zn^{2+} \rightarrow Mg^{2+} + Zn(s)[/tex].

Learn more about ions here;

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Related Questions

What is the molecular formula for C5H7 if its molar mass is 536 g/mol?

Answers

Answer:

(C5H7)8

Explanation:

It's empirical formula is given as; C5H7

Molar mass of carbon(C) = 12 g/mol

Molar mass of hydrogen(H) = 1 g/mol

We are told that it's molar mass is 536 g/mol.

To find the molecular formula;

Molecular formula = n × empirical formula

Thus;

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In order to produce CO2, a student put a sample of calcium carbonate in a flask and added hydrochloric acid using the following equation: Ca(CO3) + HCl → H2O + CO2 + CaCl2 The student finds the mass of CO2 generated to be 100 g. If the student began with 200 g of HCl and there was ample CaCO3 left unreacted, what was the percent yield?

Answers

Answer:

82.9% Is the percent yield

Explanation:

First, we need to balance the equation.

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Since the problem tells you that the excess reactant is Ca(CO3), we just need to Stoichiometry to solve for the theoretical yield.

[tex]\frac{200g HCl}{1}[/tex] x [tex]\frac{1 mole HCl}{36.46gHCl}[/tex] x [tex]\frac{1 mol CO2}{2molHCl}[/tex] x [tex]\frac{44.01CO2}{1molCO2}[/tex] =120.707 g (theoretical yield)

Take your actual yield of 100g, divide that by 120.707, then multiply that decimical by 100.

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Answers

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Answers

Explanation:

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Answer:

oxyacid, oxoacid

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Answers

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Answers

Answer:

Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil.

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Answers

Answer:

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Explanation:

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Answer:

[tex]\boxed {\boxed {\sf About \ 0.14 \ mol \ S }}[/tex]

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To convert from grams to moles we must use the molar mass. This can be found on the Periodic Table.

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[tex]\frac{ 32.07 \ g \ S }{1 \ mol \ S }[/tex]

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[tex]4.5 \ g \ S *\frac{ 32.07 \ g \ S }{1 \ mol \ S }[/tex]

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[tex]4.5 \ g \ S *\frac{1 \ mol \ S }{32.07 \ g \ S}[/tex]

[tex]4.5 *\frac{1 \ mol \ S }{32.07 }[/tex]

[tex]\frac{4.5 \ mol \ S }{32.07 }[/tex]

[tex]0.140318054 \ mol \ S[/tex]

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[tex]0.14 \ mol \ S[/tex]

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Answers

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