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3 years ago

1.Which element would most likely lose electrons to form positive ions when bonding with other elements?

1 answer:

8 0

1) The element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).

2) The correct statement about sodium atoms is; "The sodium atom transfers electrons to the chlorine atoms to form ionic bonds."

3) Based on their location in the periodic table, nitrogen (N) and oxygen (O) are most likely to form covalent bonds with each other

4) Electronegativity is best described by the phrase; "the relative strength with which an element attracts electrons in a chemical bond"

Metals of group 1 and 2 are highly electropositive and are more likely to loose electrons in a bonding situation. Therefore, the element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).

Sodium chloride is an ionic compound. It is formed by transfer of electrons from sodium to chlorine. Sodium is highly electropositive while chlorine is highly electronegative. Therefore, sodium chloride is formed when sodium atom transfers electrons to the chlorine atoms to form ionic bonds.

Covalent bonds are formed between two nonmetals. Nitrogen and oxygen are non metals hence they form covalent bonds.

According to Linus Pauling, electronegativity refers to the ability of an element in a compound to draw electrons towards itself.

Learn more: brainly.com/question/14077687

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The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago

PLS HELP THIS IS TIMED

hammer [34]

Answer:

the plant population over time has decreased. this could make all other organisms above the plant population numbers decrease as well.

Explanation:

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3 years ago

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When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0

3 years ago

4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a

yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

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3 years ago

Which of the following groups of atoms are most likely to have similar properties?

vladimir1956 [14]

The third option because the elements mentioned in that selection are in the same group, they must have similar properties.

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3 years ago

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