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2 years ago

Food deteriorates more slowly in a refrigerator because

Chemistry

2 answers:

andreev551 [17]2 years ago

6 0

The bacteria move slower and also when refrigerated there is less liquid so mold is less likely to happen.

charle [14.2K]2 years ago

5 0

Fewer collisions have the energy of activation

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Which represents the correct equilibrium constant expression for the reaction below?

Ad libitum [116K]

Answer:

$\large\boxed{\large\boxed{K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}}}$

Explanation:

1. Chemical equilibrium equation:

$Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}+2Ag(s)$

2. Species

In an equilibrium constant expression you do not include the solid substances; only gases and dissolved substances.

The symbol $(s)$ means solid, thus Cu(s) and Ag(s) shall not appear in your equilibrium constant expression.

The symbol $(aq)$ means in aqueous solution, thus the both $Ag^+$ and $Cu^{2+}$ must appear in the equilibrium constant expression.

3. Equilibrium constant expression.

It is the quotient of the product of the concentrations of the species on the right hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left hand side, each raised to its coresponding coefficient.

$K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}$

4 0

2 years ago

Read 2 more answers

Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T

GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0

2 years ago

What alkene reacts the fastest with HBr?

Rasek [7]

The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.

So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

2 years ago

Read 2 more answers

An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will

Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0

2 years ago

Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident

Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

(1) ¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

(2) N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

-(2) 2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

2×(1) H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

-(2) + (3) + 2×(1) 2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

NH₄Cl(s) → NH₃(g) + HCl(g)

Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol

I hope it helps!

8 0

2 years ago