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3 years ago

Which combination of aqueous solutions below, when mixed together, will result in evidence of a chemical reaction?

Chemistry

1 answer:

patriot [66]3 years ago

5 0

Answer:

LiBr + AgNO₃ → AgBr ↓ + LiNO₃

Option E.

Explanation:

We determine our reactants, and we analyse if there is a precipitated formed.

A. NH₄CH₃COO + AgNO₃ → AgCH₃COO + NH₄NO₃

No reaction.

Salts from acetate are slightly soluble, all the salts from nitrate are soluble

B. 2LiCl + H₂SO₄ → 2HCl + Li₂SO₄

No reaction. Sulfates can only make precipitate with Ca²⁺, Ba²⁺, Mg²⁺, Ag⁺ and Pb²⁺

C. 2NaCl + K₂SO₄ → 2KCl + Na₂SO₄

Same explanation as before. Salts from halides (Cl⁻, Br⁻, I⁻, F⁻) are mostly soluble but we have exceptions.

D. 2NaBr + Li₂CO₃ → Na₂CO₃ + 2LiBr

Salts from carbaonate and group 1 are soluble. The others elements make precipitates with carbonates.

E. Salt from bromide and silver cation is a precipitated. Bromines can also make solid when it is mixed with Hg₂²⁺, Cu⁺, Pb²⁺

E. LiBr + AgNO₃ → AgBr ↓ + LiNO₃

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For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

BlackZzzverrR [31]

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ( $\Delta G^{0}$ ) is given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

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As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.

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Answer:

See Explanation

Explanation:

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The products of electrophilic aromatic substitution depends on the substituents already present on the benzene ring. Some substituents activate the ring towards electrophilic substitution and direct the incoming electrophile to the ortho and para positions on the ring while some substituents deactivate the benzene ring towards electrophilic substitution and direct the incoming electrophlle to the meta position on the ring.

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