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Lelechka [254]
3 years ago
8

The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o

f salt in one tablespoon.
Finally, solve (remember significant figures):
Chemistry
1 answer:
finlep [7]3 years ago
3 0

Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

You might be interested in
(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
9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard
storchak [24]

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁  P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ =  6514.63 mL . atm . K  / 293 K . atm

V₂ = 22.23 mL

8 0
3 years ago
Express as an ordinary number.<br> 6.983 x 104-
Paladinen [302]

Answer:

The answer is 69,830.

Explanation:

6.983x 10^4 = 69, 830

8 0
3 years ago
How many grams of KBrO2 are there in 0.168 moles
Mnenie [13.5K]

Hey There!

Molar mass KBrO2 = 151.0011 g/mol

1 mole KBr --------- 151.0011 g

So in 0.168 moles :

mass = number of moles * molar mass

mass = 0.168 * 151.0011

mass = 25.36 g of KBrO2

5 0
3 years ago
Can someone please help me with this? thank you!
Kamila [148]

Answer:

Sulfur — S

calcium — Ca

carbon — C

silver — Ag

gold — Au

helium — He

hydrogen — H

should become easier for you to memorise them as you go on in life (meaning going up a grade… grade 7 to 8 to 9 to 10 until you reach college and whatnot)

6 0
2 years ago
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