Which chemical equation follows the law of conservation of mass?

Answers

Answer 1

Answer:

The chemical equation presented in option A follows the law of conservation of mass.

The principle of conservation of mass states, mass can neither be created nor destroyed but can be transformed from one form to another.

A reaction that follows the law of conservation of mass, must have equal number of moles each elements in reactants side and products side.

Only option A follows the law of conservation of mass;

$2LiOH \ + \ + H_2CO_3 \ ---> \ Li_2CO_3 \ + \ 2H_2O$

Thus, we can conclude that the chemical equation presented in option A follows the law of conservation of mass.

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Answer 2

Answer:

Answer:

Option A

Explanation:

The expression that obeys the law of conservation of mass is choice A;

2LiOH + H₂CO₃ → Li₂CO₃ + 2H₂O

According to the law of conservation of mass; "in a chemical reaction, matter is neither created nor destroyed". By this law, mass is usually conserved.

The equation shows that mass is conserved because the number of moles of each specie is found on both sides

Number of moles

Li O H C

Reactants 2 5 4 1

Products 2 5 4 1

This shows that mass is indeed conserved.

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Why can't 1−methylcyclohexanol be prepared from a carbonyl compound by reduction? select the single best answer?

Answers

1−methylcyclohexanol is a tertiary alcohol. Tertiary Alcohols are synthesized by either reacting Ketone with Organometallic compounds like Grignard reagent or by hydration of substituted alkenes. 1−methylcyclohexanol can not be synthesized by reduction of carbonyl compound because it is not possible to have a starting carbonyl compound having carbonyl group along with three other alkyl groups (as carbon can only form 4 bonds).

Result:
Tertiary alcohols don't contain a hydrogen atom at carbon attached to hydroxyl group that is why it is not possible to synthesize 1−methylcyclohexanol by reduction of carbonyl compound.

A spontaneous reaction occurs

Answers

Answer:

A spontaneous reaction is a reaction that occurs in a given set of conditions without intervention. A spontaneous reactions are accompanied by an increase in overall entropy, or disorder

Explanation:

Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

$2I^-(aq)---- > I_2(s)+2e^-$

Half reaction for the cell:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

$2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq)---- > I_2(s)+2e^-$ , Reduction potential is 0.53V

Reduction half reaction: $ClO_2(g)+e^----- > ClO_2^-$ ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

$2I^-(aq)---- > I_2(s)+2e^-$

Find more information about Reduction potential here:

brainly.com/question/7484965

Answer: The half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

$2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V$

Reduction half reaction: $ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V$ ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

The metal thallium becomes superconducting at temperatures below 2.39K. Calculate the temperature at which thallium becomes superconducting in degrees Celsius. Round your answer to decimal places.

Answers

Answer:

-270.76°C

Explanation:

Given that metal Thallium becomes superconducting below the temperature of 2.39 kelvin i.e. this temperature is critical temperature for Thallium and below critical temperature a metal offers no resistance to the flow of electric current. Also the metal below its critical temperature expels the magnetic field in such a way that they do not penetrate the metal and pass through its surface only.

We have the relation between kelvin scale and degree Celsius scale of temperature measurement as:

$C = K - 273.15$

$C=2.39-273.15\n C=-270.76^(o)C$

Which of the following equations violates the law of conservation of mass? A. FeCl3 + 3NaOH yields Fe(OH)3 + 3NaCl

B. CS2 + 3O2 yields CO2 + 2SO2

C. Mg(ClO3)2 yields MgCl2 + 2O2

D. Zn + H2SO4 yields H2 + ZnSO4

Answers

Hi!

The chemical equation that violates the law of conservation of mass is C

Mg(ClO₃)₂ → MgCl₂ + 2O₂

Let's see why:

The number of atoms from each element should be the same on both sides of the equation.

For Mg:1 atoms Left Side=1 atoms Right Side
For Cl:2 atoms Left Side=2 atoms Right Side
For O: 6 atoms Left Side ≠ 4 atoms Right Side

So, the equation is not balanced for O. The correct balanced equation is the following:

Mg(ClO₃)₂ → MgCl₂ + 3O₂

Have a nice day!

Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.