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3 years ago

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Answer:Electrons exhibit interference or diffraction due to scattering objects or apertures around the same size as their de Broglie wavelength. Electrons exhibit a discrete charge. Electrons exhibit a discrete mass. Electrons may tunnel through barriers of nanometer or below thickness

Explanation:

Electrons behave both as matter an waves in accordance with De Broglie principle of wave particle duality hence they possess an associated wavelength and as such undergo diffraction and interference.

Electrons have a known magnitude of mass and charge.

Many studies have also shown that electrons can tunnel through barriers of 1nm thickness or less.

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During your future endeavors in chemistry you will inevitably do a bunch of acid-base problems. Of the strong bases above, calci

Alex777 [14]

Answer:

d. There are two moles of hydroxide for each mole of these compounds.

Explanation:

calcium hydroxide, strontium hydroxide, and barium hydroxide all have two hydroxide compunds. the difference between each of these compounds is they all contain a different element along with hydroxide.

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4 years ago

Carbohydrate metabolic enzymes bind D-glucose specifically. D-glucose has an estimated caloric value of 1 k.cal per 4 grams of c

Aleksandr [31]

Answer:

D) 0 k.cal per 4 grams

Explanation:

Genarally, glucose can be classified into two enantiomers such as d-glucose and l-glucose. The d-glucose is the most common sugar that bodies of living organisms use as source of energy. However, l-glucose is an organic compound and it is one of the aldohexose monosaccharides. It is the l-isomer of glucose and commonly refer to as a low-calorie sweetener. The l-glucose is relatively indistinguishable in taste from d-glucose but cannot be used as a source of energy. Therefore, in the given problem:

if d-glucose has an estimated caloric value of 1 k.cal per 4 grams of carbohydrate, then the caloric value of l-glucose will be 0 k.cal per 4 grams.

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3 years ago

Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab

uranmaximum [27]

The answer is B) 230 m3

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3 years ago

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Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accu

AleksandrR [38]

Answer:

See explanation

Explanation:

A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.

Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.

The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.

8 0

3 years ago

Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net

Degger [83]

Answer:

- $AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

- $K=1.2x10^{-5}$

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

$AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}$

And the formation of the complex:

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7$

We obtain the balanced net ionic equation by adding the aforementioned equations:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

$K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}$

Best regards.

4 0

3 years ago