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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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Calculate the pOH of a solution that contains 3.9 x 10-5 M H3O+ at 25°C.

exis [7]

Answer:

Option b. 9.59

Explanation:

First, let us calculate the pH. This is illustrated below:

[H3O+] = 3.9 x 10-5 M

pH = —Log [H3O+]

pH = —Log [3.9 x 10-5]

pH = 4.41

Recall: pH + pOH = 14

4.41 + pOH = 14

Collect like terms

pOH = 14 — 4.41

pOH = 9.59

6 0

2 years ago

Calculate the nuclear binding energy for 5525mn in megaelectronvolts per nucleon (mev/nucleon). express your answer numerically

VARVARA [1.3K]

Answer:

8.533 Mev/nucleon

Explanation:

The given elements is:- $^{55}_{25}Mn$

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, the number of protons = 25

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

55 = 25 + Number of neutrons

Number of neutrons = 30

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,

Calculated mass = (25*1.007277 + 30*1.008665) amu = 55.441875 amu

Actual mass = 54.9380451 amu

Mass defect = Δm = |54.9380451 - 55.441875| amu = 0.5038299 amu

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

So, Energy = 0.5038299*931.5 MeV= 469.317552 MeV

Total number of nucleons in the atoms = 55

<u>So, Energy = 469.317552/55 MeV/nucleon = 8.533 Mev/nucleon</u>

5 0

2 years ago

The molar mass of an imaginary molecule is is 93.89 g/mol. Determine its density at STP.

Igoryamba

Answer:

Density = 4.191 gm/L

Explanation:

Given:

Molar mass = 93.89 g/mol

Volume(Missing) = 22.4 L (Approx)

Find:

Density at STP

Computation:

Density = Mass/Volume

Density = 93.89 / 22.4

Density = 4.191 gm/L

3 0

1 year ago

A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

bagirrra123 [75]

Answer:

The final volume of the sample of gas $V_{2}$ = 0.000151 $m^{3}$

Explanation:

Initial volume $V_{1}$ = 200 ml = 0.0002 $m^{3}$

Initial temperature $T_{1}$ = 296 K

Initial pressure $P_{1}$ = 101.3 K pa

Final temperature $T_{2}$ = 336 K

Final pressure $P_{2}$ = K pa

Relation between P , V & T is given by

$P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }$

Put all the values in the above equation we get

$101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )$

$V_{2}$ = 0.000151 $m^{3}$

This is the final volume of the sample of gas.

4 0

2 years ago

The equilibrium constant, Kc, for the reaction H2 (g) + I2 (g) ⇄ 2HI (g) at 425°C is 54.8. A reaction vessel contains 0.0890 M H

Mars2501 [29]

Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$