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Nina [5.8K]
3 years ago
10

A chemical change has occured if the number of atoms is greater after the reaction than it was before the reaction.

Chemistry
1 answer:
diamong [38]3 years ago
8 0

Answer:

False

Explanation:

It is not correct to assert that a chemical change has occurred if the number of atoms is greater after the reaction than it was before the reaction.

For every chemical reaction, the law of conservation of mass must be strictly adhered.

The law states that "during a chemical reaction, matter is neither created nor destroyed, atoms are simply rearranged".

By the virtue of this law, the number of atoms at the beginning and the end of the reaction must remain the same for this law to be applicable.

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Potassium hydrogen carbonate
MariettaO [177]

Answer:

3

Explanation:

jdidsolslwoekdldloeo we

8 0
3 years ago
The atomic mass of Cu is 63.5. Find its electrochemical equivalent​
FrozenT [24]

Answer:

The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

m = Z·I·t = Z·Q

We have;

Cu²⁺ (aq) + 2·e⁻ → Cu

Therefore, one mole of Cu, is deposited by 2 moles of electrons

The charge carried one mole of electrons = 1 Faraday = 96500 C

∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

∴ Z = 3.29015544 × 10⁻⁴ g/C = 3.29015544 × 10⁻⁷ g/C

The electrochemical equivalent of copper, Cu, is Z = 3.29015544 × 10⁻⁷ g/C

7 0
3 years ago
How many molecules are in a mole of H2O
belka [17]
1023 molecules or atoms depending on substance
7 0
3 years ago
Read 2 more answers
How many chlorine ions are required to bond with one aluminum ion and why??
antiseptic1488 [7]
3 Chlorine ions are required to bond with one aluminum ion.

In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).

Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.

Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
8 0
3 years ago
(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
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