The reaction between sodium and chlorine that forms table salt is shown

The kinetic energy within molecules of an object _ when heat is added? Increases decreases or remains the same

kotegsom [21]

The kinetic energy of an object increases when heat is added.

6 0

3 years ago

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to

valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of H₂SO₄.

Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.

To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is = 5.451 X 10³ kg.

5 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 $\sqrt 2$ a 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

rank the following list of aqueous solutions and solvents according to decreasing vapor pressure the substance with the higher v

Tpy6a [65]

Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water

Explanation:

Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.

8 0

3 years ago

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Simora [160]

Answer:- Density is 10.5 grams per mL and the substance is Silver(Ag).

Solution:- Mass of jewelry is given as 132.6 g. Initial volume of water in the graduated cylinder is 48.6 mL and the volume becomes 61.2 mL when this jewelry is submerged in the cylinder.

So, the volume of the jewelry = 61.2 mL - 48.6mL = 12.6 mL

We know that, $density=\frac{mass}{volume}$

Let's plug in the values in the above formula to calculate the density:

$density=\frac{132.6g}{12.6mL}$

$density=10.5\frac{g}{mL}$

So, the density of the jewelry is $10.5\frac{g}{mL}$ .

The substance from which the jewelry is made could easily be identified by comparing the calculated density with the density values given for different substances in the density table.

Looking at the density table, 10.5 grams per mL is the density for silver. So, the jewelry is made up from Silver.

4 0

3 years ago