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3 years ago

Neutralization Reactions

Chemistry

1 answer:

marta [7]3 years ago

4 0

Answer:

2HNO3+ Ba(OH)2 = Ba(NO3)2 + 2H2O

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + 6H2O

Explanation:

2HNO3+ Ba(OH)2 = Ba(NO3)2 + 2H2O

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + 6H2O

H+

O2-

OH-

Ba2+

Ca2+

NO3-

P 5+, 3+, 3-

H2O

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How an atom of an element reacts chemically is based on its what?

iragen [17]

How an atom reacts chemically depends on how willing it is to share electrons with others.

It’s electrons

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2 years ago

How to convert 12g of H2O to moles

Anna35 [415]

Answer:

moles = given mass/atomic mass

so H2O mass = 2 +16=18

so 12g of h2o= 12/16 = 3/4 moles

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2 years ago

a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula

Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

$M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + ( 1 \times 14 )\\\\M_e = 43\ gram/mol$

Now, n-factor is :

$n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3$

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0

2 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

3 years ago

calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?

Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

Explanation:

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as $c a^{+}$ ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to $F^{-}$ ions. This process of acceptance of electrons is termed as reduction.

3 0

3 years ago