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2 years ago

How many functional groups does the isopropanol contain that can experience this type of interaction

Chemistry

1 answer:

nalin [4]2 years ago

3 0

Answer:

This question is incomplete

Explanation:

This question is incomplete but what you should know is that isopropanol (also referred to rubbing alcohol) has just one functional group. This functional group is called the hydroxyl group (-OH) and it's the reason the compound name ends with "ol". The hydroxyl group can be seen in the structure of the compound (Isopropanol) below

H OH H

| | |

H- C - C - C - H

| | |

H H H

If there is any functional group in isopropanol required for any form of interaction, that functional group will be the hydroxyl group because that's the only functional group isopropanol has.

NOTE: Functional group is an atom or group of atoms that determines the chemical properties of a compound.

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How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?

Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

$0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}$

$\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}$

$\text{Volume of solution (in L)}=2.96L$

Therefore, the volume of solution will be 2.96 liters.

3 0

1 year ago

The data in the table below were obtained for the reaction: 2clo2 (aq) + 2 oh- (aq) --> clo3- (aq) + clo2- (aq) + h2o (l) exp

SVETLANKA909090 [29]

Lets organise the data given in the question
[ClO₂] (m) [OH⁻] (m) initial rate (m/s)
 0.060 0.030 0.0248
 0.020 0.030 0.00276
 0.020 0.090 0.00828
rate equation as follows
rate = k [ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
order with respect to ClO₂ is 2

3 0

1 year ago

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

4 0

1 year ago

6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an

xz_007 [3.2K]

Answer:

ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

Explanation:

1) Reduction potentials

First of all one should look up the reduction potentials for the species envolved:

$Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

$Fe^{3+} + e$ → $Fe^{2+}$ E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).

1) $Ce^{4+}$ reduces while $Fe^{2+}$ oxidates

(oxidation) $Fe^{2+}$ → $Fe^{3+} + e$ E°oxi=-0.771V

(reduction) $Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

(overall equation) $Fe^{2+}+Ce^{4+}$ → $Ce^{3+}+Fe^{3+}$ E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°- $\frac{RT}{nF} Ln Q$

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°= $\frac{RT}{nF} Ln K$

Isolating for 'K' gives:

$K=e^{\frac{nFE^{o} }{RT} }$

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

6 0

2 years ago

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago