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anzhelika [568]
2 years ago
8

Eugenol is a molecule that contains the phenolic functional group. Which option properly identifies the phenol in eugenol

Chemistry
1 answer:
givi [52]2 years ago
3 0

Answer:

Explanation:

Hello,

Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.

Best regards.

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Identify the true statements about surface tension. Molecules along the surface of a liquid behave differently than those in the
Alexxandr [17]

Answer:

Molecules along the surface of a liquid behave differently than those in the bulk liquid.

Cohesive forces attract the molecules of the liquid to one another.

Water forming a droplet as it falls from a faucet is a primary example of surface tension.

Explanation:

Surface tension is the force that stretches the liquid surface. This force acts normal to the surface. It is the downward force that acts on the surface of the liquids which is due to the cohesive forces of the liquids.

The water molecules are bonded by a strong hydrogen bond force which is between hydrogen atom and the electronegative oxygen atom. At the surface the water molecules are attracted strongly by other water molecules which lies below the surface and are stretched at the surface. Thus the water molecules at the surface acts differently than in the bulk liquid.

Mercury have a strong cohesive force than the water and have a higher surface tension force than the water.

Surface water acquires minimum surface area, hence acquiring spherical shape of water.

4 0
1 year ago
What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

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

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

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

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
2 years ago
A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0
2 years ago
The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t
Ira Lisetskai [31]
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
6 0
1 year ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
1 year ago
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