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liubo4ka [24]
1 year ago
8

Considering how small atoms are, what are the chances that at least one of the atoms exhaled in your first breath will be in you

r last breath?
Chemistry
1 answer:
STatiana [176]1 year ago
3 0

The answer is very probable because of how incredibly small atoms are

Very probable because of how incredibly small atoms are the chances that at least one of the atoms exhaled in your first breath will be in your last breath.

Hydrogen and oxygen always react in a 1:8 ratio by mass to form water. early investigators thought this meant that oxygen was 8 times more massive than hydrogen.

If you say that something is probable, you mean that it is likely to be true or likely to happen.

Everything around us is made up of really tiny molecules. However, such molecules are constructed from much smaller atoms. Then, even smaller protons, neutrons, and electrons are used to build those atoms. Quarks, which are even smaller particles than protons, make up protons.

The smallest unit of substance that may be disassembled without ejecting any electrically charged particles is the atom. The smallest piece of substance that displays an element's distinctive qualities is an atom. As a result, the atom serves as the basic unit of chemistry.

Learn more about smaller atoms brainly.com/question/28256098

#SPJ4

You might be interested in
(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t
11Alexandr11 [23.1K]

Answer:

a

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

b

The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

(1)

The position of an atom that fits in the octahedral site with radius \left( r \right)is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

2r=a-2R------(A)

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider \Delta {\rm{XYZ}} as follows:

(XY)^ 2 =(YZ) ^2 +(XZ)^2

Substitute XY as{\rm{R}} + 2{\rm{R + R}} and {\rm{YZ}} as a and {\rm{ZX}} as a in above equation as follows:

(R+2R+R) ^2 =a ^2 +a^ 2\\16R ^2 =2a^ 2\\ a =2\sqrt{2R}

Substitute value of aa in equation (A) as follows:

r= \frac{2\sqrt{2}R -2R }{2} \\ =\sqrt{2} -1R\\ = 0.414R

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.  

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

AD=2R+2r----(B)

   In    \Delta {\rm{ABC}},

(AB) ^2 =(AC) ^2 +(BC) ^2

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute {\rm{AC}} as a and {\rm{BC}} as a in above equation as follows:

(AB) ^2 =a ^2 +a ^2

AB= \sqrt{2a} ----(1)

Also,

AB=2R

Substitute value of 2{\rm{R}} for {\rm{AB}} in equation (1) as follows:

2R= \sqrt{2} aa = \sqrt{2} R

Therefore, the length of body diagonal is calculated using Pythagoras Theorem in \Delta {\rm{ABD}} as follows:

(AD) ^2 =(AB) ^2 +(BD)^2

Substitute {\rm{AB}} as \sqrt 2a   and {\rm{BD}} as a in above equation as follows:

(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

\sqrt 3a=2R+2r

Substitute a as \sqrt 2{\rm{R}} in above equation as follows:

( \sqrt3 )( \sqrt2 )R=2R+2r\\\\

r = \frac{2.4494R-2R}{2}\\

=0.2247R

\approx 0.225R

The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Note

An impure atom occupies the tetrahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The length of body diagonal is calculated using Pythagoras Theorem. The body diagonal is equal to the sum of the radii of two atoms. This helps in determining the relation between the radius of impure atom and radius of atom present in the unit cell.

7 0
3 years ago
What is one benefit and one limitation of comparative investigations
Ivan
<span>1. Seeing broader picture: determining gray areas, overlaps, and exclusions, biases, the primary focus.

2. Bias. Sources may have hidden agenda, personal discrimination conscience or unconscious and a limited view of the short term, long term effects.</span>
5 0
3 years ago
A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

7 0
3 years ago
Is Plasma See through?<br> A.True <br>B. False
Snezhnost [94]
A. True

you can see plasma. Or, technically, you can see the energy (light) given off by a plasma.
6 0
3 years ago
If the mercury in a nanometer raises 11.5 millimeters due to a change in pressure, what is the corresponding change in pressure
Marina CMI [18]

Answer:

kmy

Explanation:

7 0
3 years ago
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