Which of the following molecules do X and Y represent?

Identify the intermolecular forces holding the molecules together in each sample. CH3NH2 _______________________________________

omeli [17]

Answer:

CH3NH2 : Hydrogen bonding and London dispersion forces

F2 : Hydrogen bonding, dipole-dipole and London dispersion forces

NOF : Dipole-dipole forces

C2H6 : London dispersion forces

H2O : Hydrogen bonding and London dispersion forces

Increasing order of boiling point:

F2

C2H6

NOF

CH3NH2

H20

Explanation:

London dispersion forces exist in all substances

CH3NH2 : The hydrogen bonding in the amine group and the assymetry of the molecule makes the hydrogen bonding and dipole-dipole bonds from the methyl group.

F2 : The bond is non polar because its a homonuclear bond

NOF : Dipole-dipole forces due to difference in elecronegativity between all the atoms involved and assymetry

C2H6 : London dispersion forces because all dipole-dipole forces are cancelled out due to the symmetry of the molecule

H2O : Hydrogen bonding between hydrogen and oxygen and the assymetry of the molecule

Order of boiling points: London dispersion forces, followed by dipole-dipole forces and hydrogen bonding is the strongest

8 0

3 years ago

A chemist places 2.5316 g of Na 2SO 4 in a 100 mL volumetric flask and adds water to the mark. She then pipets 15 mL of the resu

Lera25 [3.4K]

Answer:

The concentration of the most dilute solution is 0.016M.

Explanation:

First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:

$[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M$

<u>First dilution</u>

We can use the dilution rule:

C₁ x V₁ = C₂ x V₂

where

Ci are the concentrations

Vi are the volumes

1 and 2 refer to initial and final state, respectively.

In the first dilution,

C₁ = 0.178 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.178M \times 15mL}{50mL} =0.053M$

<u>Second dilution</u>

C₁ = 0.053 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.053M \times 15mL}{50mL} =0.016M$

3 0

3 years ago

If an atom has 104 electrons, 157 neutrons, and 104 protons, what is its approximate atomic mass?

irina1246 [14]

Atomic mass is number of protons + number of neutrons (disregard electrons in mass calculations). 104 + 157 = 261.

8 0

3 years ago

A _____________reaction mechanism involves loss of a leaving group, formation of a ________ followed by the removal of a proton

Molodets [167]

Answer:

The correct answer is Option C (E1) and Option B (carbocation).

Explanation:

Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.

Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.

7 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago