Determine the mass of one mole of stearic acid, C 17 H 35 COOH . Show all work and calculations including units .​

10 answers
Question:

Determine the mass of one mole of stearic acid, C 17 H 35 COOH . Show all work and calculations including units .​

Answers

3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg

Taking this class as well 

1) They  are highly reactive.

2) They are highly reactive and are non metallic.

3) -2

4)LiCl

5) How electrons are shared

6) Each hydrogen electron has one valence electron

7) share two valence electrons

8) 85.0 g/mol

Explanation:

1) As both the kind of elements are ready to give or taken one electron, to attain noble gas configuration or full filled stability, they are highly reactive.

The element which has just one electron in valence shell will give the electron easily.

The element which has seven valence electrons will accept one electron easily.

2) Halogen have seven valence electrons. They are highly reactive as the are ready to take one electron. As they cannot give electrons they are non metals.

3) Generally oxygen can accept two electrons so its charge is "-2"

4) Formula unit is generally for ionic compound. LiCl is an ionic compound.

5) In lewis structure we write elements are then distribute electrons over all the atoms. It shows the number of bonded electrons, unshared or lone electrons as well.

6) The configuration of hydrogen is 1s1, thus it has only one valence electron/ atom.

7) In each covalent bond there are two shared electrons (shared by two atoms bonded).

8) molar mass = atomic mass of Na + Atomic mass of N + 3 X atomic mass of O

Molar mass = 23 + 14 + 3(16) = 85.0 g/mol

For 3: The total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]

For 4: The mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]

For 5: The equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]

Explanation:

For 3:

To calculate the mass change of the reaction for given energy released, we use Einstein's equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = [tex]3.83\times 10^{-12}J[/tex]

[tex]\Delta m[/tex] = mass change = ?

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]3.83\times 10^{-12}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=4.255\times 10^3kg[/tex]

Hence, the total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]

For 4:

For the given isotopic representation:  [tex]_{27}^{60}\textrm{Co}[/tex]

Atomic number = Number of protons = 27

Mass number = 60

Number of neutrons = Mass number - Atomic number = 60 - 27 = 33

To calculate the mass defect of the nucleus, we use the equation:

[tex]\Delta m=[(n_p\times m_p)+(n_n\times m_n)+]-M[/tex]

where,

[tex]n_p[/tex] = number of protons  = 27

[tex]m_p[/tex] = mass of one proton  = 1.00728 amu

[tex]n_n[/tex] = number of neutrons  = 33

[tex]m_n[/tex] = mass of one neutron = 1.00867 amu

M = Nuclear mass number = 59.9338 amu

Putting values in above equation, we get:

[tex]\Delta m=[(27\times 1.00728)+(33\times 1.00867)]-[59.9338]\\\\\Delta m=0.54887amu[/tex]

Converting the value of amu into kilograms, we use the conversion factor:

[tex]1amu=1.66\times 10^{-27}kg[/tex]

So, [tex]0.54887amu=0.54887\times 1.66\times 10^{-27}kg=0.911\times 10^{-27}kg[/tex]

To calculate the equivalent energy, we use the equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = ?

[tex]\Delta m[/tex] = mass change = [tex]0.911\times 10^{-27}kg[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]E=(0.911\times 10^{-27}kg)\times (3\times 10^8m/s)^2\\\\E=8.199\times 10^{-11}J[/tex]

Converting this into kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, [tex]8.199\times 10^{-11}J=8.199\times 10^{-14}kJ[/tex]

Hence, the mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]

For 5:

For the given chemical reaction:

[tex]C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l);\Delta H=-1418kJ/mol[/tex]

To calculate the equivalent mass of the reaction for given energy released, we use Einstein's equation:

[tex]E=\Delta mc^2[/tex]

E = Energy released = [tex]1418kJ=1418\times 10^3J[/tex]

[tex]\Delta m[/tex] = mass change = ?

c = speed of light = [tex]3\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]1418\times 10^{3}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=1.5755\times 10^{-11}kg[/tex]

Hence, the equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]

Answers explained below

Explanation:

(a)Given,

Molecular formula of the complex = [Ni(NH3)x(H2O)y]Clz

(i) Ni is in +2 oxidation state in the complex.

(ii) NH3 and H2O are the neutral ligands but Cl  is the negatively charged ligand.

(iii) complex is neutral

So, to make the nickel complex in +2 oxidation with neutral charge, we requires 2 Cl-.

Hence, form the above statements, we can say that here in the complex

z=2

(b) Molarity of HCl = 0.2005M

Volume of HCl used = 20.02mL = 20.02*10-3 L

Weight of the nickel(II) ammonia complex = 0.1550g

Reaction of HCl with Ammonia,

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl reacts with ammonia in 1:1 ratio to form ammonium salt (NH4Cl). That means 1 mol of HCl reacts with 1 mol of NH3.

So, we have to find number of moles of HCl used.

No. of moles of HCl used = Molarity of HCl * Volume of HCl used (L)

= 0.2005M * 20.02*10-3 L = 4.014*10-3 moles

Hence no. of moles of ammonia in the complex = No. of moles of HCl used = 4.014*10-3 mol

So, Experimental Equivalent Weight = Weight of the nickel(II) ammonia complex/ No. of moles of NH3

= 0.1550g / 4.014*10-3 mol

= 38.615 g/mol

Hence, Experimental Equivalent Weight = 38.615 g/mol

(c) Given,

x+y=< 6

Molar mass of [Ni(NH3)x(H2O)y]Clz = 58.69 + x(14.00+3*1) + y(16+2*1)+z(35.45)

= (58.69 + 17x + 18y + 35.45z) g/mol

Case1 x=6, y=0 and z=2

Molar mass of [Ni(NH3)6(H2O)0]Cl2   = (58.69 + 17*6 + 18*0 + 35.45*2) g/mol

= 58.69+102+70.90 = 231.59 g/mol

Experimental Equivalent Weight = 233.59/6 = 38.598 g/mol

So, This experimental equivalent weight is equal to the calculated experimental equivalent weight.

Hence the molecular formula of the complex is [Ni(NH3)6]Cl2 where x=6, y=0 and z=2.

Note: You can try other combination but in every case you will find lower or higher calculated experimental equivalent weight.

Nitrogen (N) has a molar mass of about 14.007 g/mol, so you have

100.0 g N₂ = (100.0 g) • (1/28.014 mol/g) ≈ 3.570 mol N₂

Hydrogen (H) has a molar mass of about 1.008 g/mol, so

100.0 g H₂ = (100.0 g) • (1/2.016 mol/g) ≈ 49.60 mol H₂

In the balanced reaction, 1 mole of N₂ reacts with 3 moles of H₂ to make 2 moles of NH₃. We have

3.570 • 3 ≈ 10.71

so the reaction would use up all 3.570 mol N₂ and 10.71 mol H₂ to produce about 7.139 mol NH₃.

(A) H₂ is the excess reactant. There is an excess of 49.60 - 10.71 ≈ 38.89 mol H₂.

(B) N₂ is the limiting reactant. All of the N₂ gets consumed in the reaction.

(C) NH₃ has a molar mass of (14.007 + 3 • 1.008) g/mol = 17.031 g/mol. This reaction would theoretically yield about 7.139 mol NH₃, or

7.139 mol NH₃ = (7.139 mol) • (17.031 g/mol) ≈ 121.6 g NH₃

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Explanation:

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E= m× C2
... than:
m=E/c 2
m=1,366•10^3/3•10 A 8 2
m=1,518•10^14 kg

Q1. They are highly reactive. Q2. High reactivity, nonmetallic. Q3. Oxygen has an ion charge of -2. Q4. LiCl I believe. Q5. How electrons are shared. Q6 1. Q7. Share 2 valence electrons, I believe.

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