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Free_Kalibri [48]
3 years ago
6

Borane (BH3) adds to alkenes to form an alkylborane. In the first box draw the mechanism arrows, and in the second box draw the

correct product. Be sure to add lone pairs of electrons and nonzero formal charges to all species.

Chemistry
1 answer:
OLEGan [10]3 years ago
6 0

Answer:

The reaction when the Borane (BH3) is add to an alkene and form an alkylborane is shown below.

Explanation:

The boron of the borane does not have extra electron pairs, in this way the double bond of the alkene attacks the boron and the hydrogen belonging to the borane adheres to the carbon that is more substituted, thus forming an alkyl borane.

You might be interested in
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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#SPJ1

7 0
2 years ago
The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.
yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

Q_{1} =3.25*(-40.7)=-132.275kJ

The work done:

W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ

The change in internal energy:

delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ

7 0
3 years ago
Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when di
Gwar [14]

Answer:

NaCl neutral

Na₂SO₄ basic

NaHCO₃ basic

K₃PO₄ basic

Na₃C₆H₅O₇ basic

CaCl₂ neutral

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water.

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water

Explanation:

<em>Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs.</em>

<em />

NaCl

When NaCl is dissolved in water the resulting solution is neutral.

Na₂SO₄

When Na₂SO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the SO₄²⁻ ion.

SO₄²⁻(aq) + H₂O(l) ⇄ HSO₄⁻(aq) + OH⁻(aq)

NaHCO₃

When NaHCO₃ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the HCO₃⁻ ion.

HCO₃⁻(aq) + H₂O(l) ⇄ H₂CO₃(aq) + OH⁻(aq)

K₃PO₄

When K₃PO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the PO₄³⁻ ion.

PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

Na₃C₆H₅O₇ (sodium citrate)

When Na₃C₆H₅O₇ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the C₆H₅O₇³⁻ ion.

C₆H₅O₇³⁻(aq) + H₂O(l) ⇄ C₆H₆O₇²⁻(aq) + OH⁻(aq)

CaCl₂

When CaCl₂ is dissolved in water the resulting solution is neutral.

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water according to the following equation:

Na₂C₂O₄(aq) ⇄ 2 Na⁺(aq) + C₂O₄²⁻(aq)

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water according to the following equation:

C₂H₂O₄(aq) ⇄ C₂HO₄⁻(aq) + H⁺(aq)

4 0
3 years ago
Find the fugacity coefficient of a gaseous species of mole fraction 0.4 and fugacity 25 psia in a mixture at a total pressure of
Oksana_A [137]

<u>Answer:</u> The fugacity coefficient of a gaseous species is 1.25

<u>Explanation:</u>

Fugacity coefficient is defined as the ratio of fugacity and the partial pressure of the gas. It is expressed as \bar{\phi}

Mathematically,

\bar{\phi}_i=\frac{\bar{f_i}}{p_i}

Partial pressure of the gas is expressed as:

p_i=\chi_iP

Putting this expression is above equation, we get:

\bar{\phi}_i=\frac{\bar{f_i}}{\chi_iP}

where,

\bar{\phi}_i = fugacity coefficient of the gas

\bar{f_i} = fugacity of the gas = 25 psia

\chi_i = mole fraction of the gas = 0.4

P = total pressure = 50 psia

Putting values in above equation, we get:

\bar{\phi}_i=\frac{25}{0.4\times 50}\\\\\bar{\phi}_i=1.25

Hence, the fugacity coefficient of a gaseous species is 1.25

8 0
3 years ago
Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti
geniusboy [140]
1) The forward reaction is N2 (g) + O2 (g)  → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

  N2 (g) + O2 (g)  ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)
4 0
3 years ago
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