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tigry1 [53]
3 years ago
15

Identify the Lewis acid in the following reaction: Cr3+(aq)+6H2O(aq)⇌Cr(H2O)63+(aq)

Chemistry
2 answers:
Mashcka [7]3 years ago
6 0

Answer:The Lewis acid the in the given reaction is Cr^{3+} ion.

Explanation:

According Lewis Theory of  acid base concept:

  • Acids are those who accepts pair of electron pair
  • Bases are those who donates pair of electron pair.

Cr^{3+}(aq)+6H_2O(aq)\rightleftharpoons Cr(H2O)_{6}^{3+}(aq)

In the above reaction chromium (III) ion is acting as a Lewis acid and water is acting as Lewis base. Chromium ion easily accepts the electron pair that is lone pair of oxygen atom present in water molecule.

Nostrana [21]3 years ago
3 0
In the reaction,
Cr3+(aq) + 6H2O -------> [Cr(H2O)6]3+(aq)
Cr3+ IS THE LEWIS ACID H2O is the Lewis base.
A Lewis acid is a compound or chemical that accepts a lone pair of electrons. In the above equation, Cr3+ is accepting the lone pair of electrons.
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= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl

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hurry please! avogadro's law relates the volume of a gas to the number of moles of gas when temperature and pressure are constan
DIA [1.3K]

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

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V1/n1 = V2/n2

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5 0
3 years ago
A certain system absorbs 350 joules of heat and has 230 joules of work done on it what is the value of
marysya [2.9K]
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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