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4 years ago

What are the names of the compounds KCIO3 and H2SO4 respectively

Chemistry

2 answers:

Andre45 [30]4 years ago

5 0

Answer:

Then name of the compound KCl $O_{3}$ is Potassium Chlorate and $H_{2}$ S $O_{4}$ is Sulfuric Acid.

Explanation:

Here our first compound is KCl $O_{3}$ which contains potassium, chlorine and oxygen atoms. It is a white crystalline substance.

$H_{2}$ S $O_{4}$ is Sulfuric acid which contains sulfur, oxygen and hydrogen. It is also known as Vitriol.

aksik [14]4 years ago

4 0

Answer:

Explanation:

Hi friend, Potassium chlorate is a a white crystalline compound containing potassium, chlorine and oxygen atoms, with the molecular formula KClO3.Jul 26, 2017

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3 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

4 years ago

What are the names of the compounds KCIO3 and H2SO4 respectively​

What are the names of the compounds KCIO3 and H2SO4 respectively