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2 years ago

How many moles of carbonate are there in sodium carbonate

Chemistry

2 answers:

vaieri [72.5K]2 years ago

5 0

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>

The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

= 46 + 12 + 48 = 106g/mol

mass of CO3 = 12 + 48 = 60g

no. of moles of CO3 = 60/106

no. of moles of CO3 = 0.566mol

Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

ivolga24 [154]2 years ago

4 0

<h2>Answer:</h2>

molecular weight of Sodium Carbonate or mol The molecular formula for Sodium Carbonate is Na2CO3. The SI base unit for amount of substance is the mole.','.

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ollegr [7]

Answer:

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3 years ago

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(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:

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

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.

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3 years ago

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat

Valentin [98]

Answer:

(a)

$0.0342M$

(b)

$t_{1/2}=17.36s\\t_{1/2}=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\$

$[NOBr]=\frac{1}{29.2/M}=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_{1/2}=\frac{1}{k[NOBr]_0}$

Therefore, for the given initial concentrations one obtains:

$t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s$

Best regards.

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3 years ago

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Answer:

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3 years ago

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How many moles of carbonate are there in sodium carbonate​

How many moles of carbonate are there in sodium carbonate