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3 years ago

A class of substances known as __________ are used for their medicinal properties.

Chemistry

2 answers:

salantis [7]3 years ago

8 0

Answer:

Alkaloids

Explanation:

Alkaloids are organic compounds that are rich in nitrogen which are commonly found in plants though some may be prepared artificially.

Alkaloids helps in protecting plants against pathogens and animals. However they could also be useful in human and veterinary medicine. They have various pharmacological effects. Example of alkaloids include:

Morphine:They can be used to cure and relief pain

Quinine:fight malaria infection

Chelerythrine: fight bacteria .

All these are found in various plants.

Misha Larkins [42]3 years ago

6 0

Answer: Alkaloids

A class of substances known as alkaloids are used for their medicinal properties.

Explanation:

Alkaloids are small but complex organic substances with at least one nitrogen atom in its ring structure. Hence, they have strong basic properties, and are produced in naturally by some plants.

Examples of alkaloids and their medicinal properties are as follows:

- caffeine, used in certain drug and drinks to stimulate the nervous system

- morphine, used to reduce pain and induce sleep in humans.

- cocaine, nicotine etc

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Crude oil may contain hundreds of different types of hydrocarbons. Some examples include: Butane(C4H10) Dodecane (C12H26) Octane

Nesterboy [21]

Answer:

Butane and benzene are both compounds. They contain multiple elements bonded together in a specific ratio. Kerosene and gasoline are mixtures because they are combinations of several compounds. Oxygen is an element because it is made up of only one type of atom.

Explanation:

7 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

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3 years ago

Hey I have a chemistry test tomorrow,any good test advice or tips for me?

Vikentia [17]

Study no other tips needed.

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3 years ago

Read 2 more answers

A balloon is filled with 0.250 mole of air at 35Â°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure

lord [1]

Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).

n = 0.250 mol; amount of substance.

V = 6.23 L; volume of the balloon.

T = 35°C = 308.15 K; temperature.

R = 0.08206 L·atm/mol·K, universal gas constant.

Ideal gas law: p·V = n·R·T.

p = n·R·T / V.

p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.

p = 1.015 atm; presure of the air.

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