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2 years ago

Which orbitals form a pi bond?

Chemistry

1 answer:

bekas [8.4K]2 years ago

8 0

Answer:

p and d orbitals

Explanation:

A π bond forms when two orbitals overlap side-on.

The most common types are formed by the overlap of p orbitals (Fig. 1).

However, d orbitals can also overlap sideways with each other to form dπ-dπ bonds and with p orbitals to form dπ-pπ bonds (Fig.2).

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Which will get spoiled faster: plum in water or plum in the open?

lara31 [8.8K]

Answer:

It depends

(plum will spoil more quickly at warm temperatures)

Explanation:

3 0

2 years ago

Balance the following equation:

Firdavs [7]

Answer: a) $2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

b) 1 mole of $SO_2$ is produced.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The skeletal equation is:

$K_2CrO_4+Na_2SO_3+HCl\rightarrow KCl+Na_2SO_4+CrCl_3 +H_2O$

The balanced equation will be:

$2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.

b) Oxidation: $2I-^-\rightarrow I_2+2e-^-$

Reduction: $SO_4^{2-}+2e^-+4H^+\rightarrow SO_2+2H_2O$

Net reaction: $2I-^-+SO_4^{2-}+4H^+\rightarrow I_2+SO_2+2H_2O$

When 1 mole of $I_2$ is produced, 1 mole of $SO_2$ is produced.

8 0

3 years ago

How does a field experiment differ from a natural experiment?

Marina CMI [18]

Answer : behavior in a field experiment is more likely to reflect real life because of its natural setting, i.e. higher ecological validity than a lab experiment

Explanation:

8 0

2 years ago

Cytochromes are critical participants in the electron transport chains used in photosynthesis and cellular respiration. How do c

goblinko [34]

Answer:

4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.

Explanation:

The cytochromes are proteins that contain heme prosthetic groups. Cytochromes undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome:

$Cytochrome-Fe²⁺ ⇄ cytochrome-Fe³⁺-e⁻$

The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes b and c₁, and a Fe-S center. This second complex reduces cytochrome c, a water-soluble membrane peripheral protein. Cytochrome c, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes a, a₃ and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.

Each transporter "downstream" is more electronegative than its neighbor "upstream"; oxygen is located in the inferior part of the chain. Thus, the electrons fall in an energetic gradient in the electron chain transport to a more stable localization in the electronegative oxygen atom.

7 0

2 years ago

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

6 0

3 years ago

Read 2 more answers