What is the difference between a jar of hydrogen and oxygen and a jar of water vapor

A solution is made by dissolving

nikklg [1K]

Answer:

THE MOLARITY IS 2.22 MOL/DM3

Explanation:

The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water

Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol

Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;

37.5 g of Na2S = 217 g of water

( 37.5 * 1000 / 217 ) g = 1000 g of water

So, 172.81 g/dm3 of the solution

So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass

Molarity = 172.81 g/dm3 / 78 g/mol

Molarity = 2.22 mol/dm3

The molarity of the solution is 2.22 mol/dm3

4 0

4 years ago

Aryan want to deposit silver on an iron spoon using silver nitrate as electrolyte. Which terminal of the battery he should conne

34kurt

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

<h3>What is electroplating?</h3>

Electroplating is a way of electrolysis, where a thin layer of metal is used to plate a determined object. It is a kinda process to purify a material you want.

The anode contains the metal you want to plate with, in this case, the Ag.

The anode contains the half-reaction of oxidation so:

Ag(s) → Ag ⁺ (aq) + e⁻

In the cathode, you have the spoon, which it takes place the half-reaction of reduction:

Ag ⁺ (aq) + e⁻ → Ag(s)

The electrolytic cell, where the redox reaction takes place, must be filled with a AgNO₃ solution.

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

Learn more about the electroplating here:

brainly.com/question/20112817

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8 0

2 years ago

Why are valence electrons important?

Usimov [2.4K]

They are the outer layer of the electron layers.

5 0

3 years ago

Read 2 more answers

(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

4 years ago

How many pounds of manganese are in 1.261 tons of Mn2O3?

Alina [70]

Answer:

2000pound

Explanation:

Manganese metal is produced from the manganese(III) oxide, Mn2O3, which is found in manganite, a manganese ore. The manganese is reduced from its +3 oxidation state in Mn2O3 to the zero oxidation state of the uncharged metal by reacting the Mn2O3 with a reducing agent such as aluminum or carbon. How many pounds of manganese are in 1.261 tons of Mn2O3? (1 ton = 2000 pounds)

About 40 different substances called organophosphorus compounds are registered in the United States as insecticides. They are considered less damaging to the environment than some other insecticides because they breakdown relatively rapidly in the environment. The first of these organophosphorus insecticides to be produced was tetraethyl pyrophosphate, TEPP, which is 33.11% carbon, 6.95% hydrogen, 38.59% oxygen, and 21.35% phosphorus. It has a molecular mass of 290.190.

4 0

2 years ago