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2 years ago

Explain why vegetables last longer in a fridge.

Chemistry

2 answers:

jeka57 [31]2 years ago

6 0

It has to do with the releasing of ethylene, which speeds up the ripening process

viva [34]2 years ago

5 0

We refrigerate nourishments to keep microscopic organisms, yeast, and moles from the great temperature they have to develop. the dampness control accessible in numerous fridges likewise eases back the weakening of nourishments, with the goal that two of the three great circumstances for the microorganisms development are wiped out. despite the fact that the microorganism develop is eased back down at low temperatures, it despite everything can happen at the 38 degrees of a customary cooler. thus, the form that develops on overlooked extras in the rear of a fridge.

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Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

Answer: The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

Explanation:

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

What type of chemical reaction is h2so4 + 2kohk2so4 + 2h2o

olchik [2.2K]

This is a neutralization reaction.

5 0

3 years ago

How did the masses of objects affect the strength of gravity

omeli [17]

Answer:

the strength of gravitational force between two objects depends on two factors,mass and distance. the force of gravity the masses exert on each other. if one of the masses is doubled, the Force of gravity between the object is doubled. increases the force of gravity decreases.

5 0

2 years ago

21-B. Mn was used as an internal standard for measuring Fe by atomic absorption. A standard mixture containing 2.00 mg Mn/mL and

Korvikt [17]

Answer:

0.0693M Fe

Explanation:

It is possible to quantify Fe in a sample using Mn as internal standard using response factor formula:

F = A(analyte)×C(std) / A(std)×C(analyte) (1)

Where A is area of analyte and std, and C is concentration.

Replacing with first values:

F = 1.05×2.00mg/mL / 1.00×2.50mg/mL

F = 0.84

In the unknown solution, concentration of Mn is:

13.5mg/mL × (1.00mL/6.00mL) = 2.25 mg Mn/mL

Replacing in (1) with absorbances values and F value:

0.84 = 0.185×2.25mg/mL / 0.128×C(analyte)

C(analyte) = 3.87 mg Fe / mL

As molarity is moles of solute (Fe) per liter of solution:

$\frac{3.87 mg Fe}{mL} \frac{1g}{1000mg} \frac{1mol}{55.845g} \frac{1000mL}{1L}$ = 0.0693M Fe

7 0

3 years ago

Read 2 more answers

Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la

tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

$\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}$

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

$-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]$

2. Derive the rate law

Assume k₋₁ ≫ k₂.

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is

rate = k[A][B] where k = k₂K

5 0

3 years ago