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soldi70 [24.7K]
3 years ago
14

Substances that are likely to disassociate in water

Chemistry
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

sugar,dirt,spit

Explanation:

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A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of
horrorfan [7]
The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s) 

Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol

Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429 

Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,

Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
5 0
3 years ago
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
The equation most often associated with Newton's Second Law of Motion is
nignag [31]

Answer:

F = ma or Force = mass x acceleration.

Explanation:

Newtons second law is about the force put onto an object. To find this you would use the formula:

Force is equal to mass times acceleration (F=ma)

7 0
3 years ago
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What type of flower is this? This is for a science project.
kenny6666 [7]

I believe that is a plant called Wisteria.

4 0
3 years ago
Read 2 more answers
Use the solubility generalizations on the information page to predict if one or more precipitates will form when aqueous solutio
neonofarm [45]

Answer:

3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)

Explanation:

In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>

The reaction is:

<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>
6 0
2 years ago
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