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3 years ago

What is orbital theory of bonding in metals

1 answer:

5 0

Metals of Group 1 donate 1 electron from its ns orbital to form ionic bond, where n is the no. of its outermost shell.
Metals of Group 2<span> donate 2 electrons from its ns orbital to form ionic bond, where n is the no. </span>of its <span>outermost shell. </span>

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Which term names elements that are usually shiny and good conductors of heat and electricity?

Sonbull [250]

Elements that are usually shiny and good conductors of heat and electricity are metals.

<h3>What are conductors?</h3>

The conductors are the metal or non metal elements which allow heat and electricity to easily pass through them.

Metals are elements with good conductivity of electric current and heat. Metals have lustre characteristics. They are shiny and bendable just like copper wire, which is a metal.

Hence, elements that are usually shiny and good conductors of heat and electricity are metals.

Learn more about conductors.

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4 0

2 years ago

A student measures the mass of a 10m^3 block of gold to be 500 kg what is the density of the gold

Anuta_ua [19.1K]

Answer:

d = 50 kg/m³

Explanation:

density = $\frac{mass}{volume}$

mass = 500 kg

volume = 10m³

$\frac{500}{10}$ = 50

d = 50 kg/m³

8 0

3 years ago

1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the

borishaifa [10]

A. 1
B. 4
C. 5
D. 3
E. 2

The correct answer is E. 2

8 0

2 years ago

Does the law of inertia apply to both moving and non-moving objects

zhuklara [117]

Yes becuase it says it maintains its velocity if it is moving or at rest

8 0

3 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>

6 0

3 years ago