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lorasvet [3.4K]
2 years ago
8

4 Na + O2 → 2 Na2O 6.79 moles of O2 will react to form how many moles of Na2O?

Chemistry
1 answer:
Nataly_w [17]2 years ago
6 0

Answer:

13.94moles of Na₂O

Explanation:

The balanced reaction expression is given as:

        4Na  +  O₂  →   2Na₂O

Given parameters:

Number of moles of O₂ = 6.97moles

Unknown:

Number of moles of Na₂O

Solution:

 To solve this problem;

            1 mole of O₂  will produce 2 moles of Na₂O ;

            6.97 moles of O₂ will produce 6.97 x 2  = 13.94moles of Na₂O

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What is the redox half equation for 3Ag2S + 2Al --> 6Ag + Al2, and identity which material is oxidized and which is reduced?
marta [7]

Answer:

Al is oxidized while Ag is reduced.

Explanation:

The complete molecular equation is;

3Ag2S + 2Al --> 6Ag + Al2S3

Oxidation half equation;

2Al ------> 2Al^3+ + 6e

Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

2Al + 6Ag^+ ----->2Al^3+ + 6Ag

Hence; Al is oxidized while Ag is reduced.

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2 years ago
Covert to scientific notation.
gtnhenbr [62]

Answer:

19) 3.6920 × 10⁻⁵

20) 4.059× 10²

21) 2.46810121416 × 10³

22)  1.0000× 10⁻³

Explanation:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

For example:

0.000036920

In scientific notation =  3.6920 × 10⁻⁵

405.9

In scientific notation  = 4.059× 10²

2468.10121416

In scientific notation  = 2.46810121416 × 10³

0.0010000

In scientific notation  = 1.0000× 10⁻³

8 0
3 years ago
Magnesium hydroxide reacts with hydrogen chloride, a double-displacement reaction occurs and produces magnesium chloride and wat
Natalija [7]

Explanation:

magnesium Hydroxide + Hydrochloric react together and give us magnesium chloride + water

5 0
2 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.

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Answer:

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Explanation:

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