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expeople1 [14]
3 years ago
7

The effective nuclear charge is the amount of charge experienced by an electron taking into account any shielding effects from o

ther electrons. Electrons between the nucleus and the electron of interest cause shielding and reduce the actual charge felt by the electron of interest. The outermost electron of boron experiences a lower effective nuclear charge than carbon. Why?
Chemistry
1 answer:
Agata [3.3K]3 years ago
6 0

Answer:

The nuclear charge increases from boron to carbon, but there is no additional shielding( that is no additional shells).

Explanation:

First of all, we must know the electron configuration of carbon and boron.

Boron- 1s2 2s2 2p1

Carbon- 1s2 2s2 2p2

Moving from boron to carbon, the effective nuclear charge increases without a corresponding increase in the number of shells. Remember that shielding increases with increase in the number of intervening shells between the outermost electron and the nucleus. Since there isn't an increase in shells, boron experience a lower screening effect.

From

Zeff= Z- S

The Z for carbon is 6 while for boron is 5 even though both have the same number of screening electron S(4 screening electrons). Hence it is expected the Zeff(effective nuclear charge) for boron will be less than that of carbon.

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For the reaction below, complete the rate expression that relates the change in concentration with respect to time to the rate o
Ann [662]

Answer: Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of Cl_2= -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = \frac{1d[NOCl]}{2dt}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO+Cl_2\rightarrow 2NOCl

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of  = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of = -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = +\frac{1d[NOCl]}{2dt}

5 0
3 years ago
1. If the atomic number of an element is 6 and its mass number is 14, how many neutrons are contained in the nucleus?
Iteru [2.4K]

You can find the neutrons by subtracting the mass by the atomic number. 14-6=8 neutrons

8 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
Can someone help with question 6 ASAP
Lera25 [3.4K]

Answer:

A metalloid is a type of chemical element which has a preponderance of properties in between, or that are a mixture of, those of metals and nonmetals. There is no standard definition of a metalloid and no complete agreement on which elements are metalloids. Despite the lack of specificity, the term remains in use in the literature of chemistry.

A series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements—boron and silicon—in more detail.

Explanation:

i hope this helps you :)

3 0
2 years ago
5. A fireplace fire heating a room on a cold day is an example of what type of heat
Inessa [10]

Answer:

I think the answer is C- radiation

Explanation:

If its not C its A Hope this helps forgive me if i'm wrong :/

8 0
2 years ago
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