Explanation: Water is able to dissolve many substances because of its polar molecules, The positive and negative ends of a water molecule attracts the molecules of other polar substances..

I'm pretty sure, idk college stuff

3 0

3 years ago

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Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

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

<u>Explanation:</u>

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

How many grams of solid barium sulfate form when 30.0 mL of 0.160 M barium chloride reacts with 70.0 mL of 0.065 M sodium sulfat

Advocard [28]

Answer:

1.06g of BaSO₄ is produced

Explanation:

BaCl₂(aq) + Na₂SO₄(aq) ⇄ 2NaCl(aq) + BaSO₄(aq)

mole BaCl₂ = (0.16M × 0.030L) = 0.0048mole BaCl₂

mole Na₂SO₄ = (0.065M × 0.070L) = 0.00455mole Na₂SO₄

Amount of product BaSO₄ produce from each reagent

BaCl₂ = (0.0048 BaCl₂) × ( 1 mol BaSO₄ / 1mol BaCl₂) × (233.3896g BaSO₄ / 1mol BaSO₄)

= 1.12g BaSO₄

Na₂SO₄ = (0.00455 Na₂SO₄) × ( 1 mol BaSO₄ / 1mol Na₂SO₄) × (233.3896g BaSO₄ / 1mol BaSO₄)

= 1.06g BaSO₄

4 0

3 years ago

Which one is correct Lewis diagram for CO2​

Which one is correct Lewis diagram for CO2