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Goshia [24]
3 years ago
14

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

ns in each reaction. Cr2(SO4)3(aq)+(NH4)2CO3(aq)−→ Cr2(SO4)3(aq)+(NH4)2CO3(aq)→ Ba(NO3)2(aq)+K2SO4(aq)−→ Ba(NO3)2(aq)+K2SO4(aq)→ Fe(NO3)2(aq)+KOH(aq)−→
Chemistry
1 answer:
Dmitrij [34]3 years ago
3 0

Answer :

(a) The net ionic equation will be,

2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)

The spectator ions are, NH_4^{+}\text{ and }SO_4^{2-}

(b) The net ionic equation will be,

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

The spectator ions are, K^{+}\text{ and }NO_3^{-}

(c) The net ionic equation will be,

Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)

The spectator ions are, K^{+}\text{ and }NO_3^{-}

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)

The ionic equation in separated aqueous solution will be,

2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)

In this equation, NH_4^{+}\text{ and }SO_4^{2-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)

(b) The given balanced ionic equation will be,

Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)

The ionic equation in separated aqueous solution will be,

Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)

In this equation, K^{+}\text{ and }NO_3^{-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

(c) The given balanced ionic equation will be,

Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)

The ionic equation in separated aqueous solution will be,

Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)

In this equation, K^{+}\text{ and }NO_3^{-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)

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Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

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2 years ago
What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?
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Answer:

  • Volume = <u>2.0 liter</u> of 1.5 M solution of KOH

Explanation:

<u>1) Data:</u>

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

<u>2) Formula:</u>

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

  • M = n / V in liter

<u>3) Calculations:</u>

  • Solve for n: M = n / V ⇒ V = n / M

  • Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

You must use 2 significant figures in your answer: <u>2.0 liter.</u>

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What is the pH of a solution with a H+ ion of 2×10-¹²?​
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Answer:

11.7

Explanation:

The pH is the negative logarithm of the concentration of H+ ions. If the concentration is 2×10-¹² the the pH will be -log(2×10-¹²) which is 11.698 which can be round up to 11.7.

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How many molecules in 9.18 moles of C11H12O22
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Answer:

55.3 × 10²³ molecules

Explanation:

Given data:

Number of moles of C₁₁H₁₂O₂₂ = 9.18 mol

Number of molecules = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

For given data:

9.18 mol × 6.022 × 10²³ molecules /1 mol

55.3 × 10²³ molecules

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