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FromTheMoon [43]

3 years ago

6

Lipids (also known as fats) have two types of monomers called

1 answer:

dalvyx [7]3 years ago

4 0

Fats are large molecules made of two types of molecules, glycerol and some type of fatty acid.

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What is the atomic mass, in amu, of this atom?

hjlf

Answer:precisely 1/12 the mass of an atom of carbon-12.

Explanation:The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus. In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass.

7 0

3 years ago

When gases are compressed, the particles are forced closer together.

Elanso [62]

Gases can be compressed easily because there is a large amount of space between the individual molecules, which are very active and move around at high speed. When gases are compressed, the particles are forced much closer together, allowing a huge amount of particles to fit a small space.

5 0

3 years ago

Calculate the net force on the object below.

Anastaziya [24]

Answer:

first we add the same direction. 12N + 32 N=44N .

then we add the forces. 54 up + 44N down= 10N up

8 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

4 years ago

Calculate the molarity of 0.50 mol MgCl2 in 750 ml of solution.

Vedmedyk [2.9K]

Answer : The correct answer for molarity of MgCl₂ solution is 0.67 M.

Molarity :

It expresses concentration of any solution . It can be defined as mole of solute present in volume of solution in Liter .

It uses unit M or $\frac{mol}{L}$

It can be formulated as :

$Molarity (\frac{mol}{L} ) = \frac{mole of solute (mol)}{volume of solution (L)}$

Given : mol of MgCl₂ (solute ) = 0.50 mol

Volume of solution = 750 mL

Following steps can be done :

Step 1 ) To convert volume of solution from mL to L

1 L = 1000 mL

$Volume of solution = 750 mL * \frac{1 L}{1000mL}$

Volume of solution = 0.750 L

Step 2) To find molarity

Plugging mole of solute and volume of solution in molarity formula as :

$Molarity = \frac{0.50 mol}{0.750 L }$

$Molarity = 0.67 \frac{mol}{L}$

Hence molarity of MgCl₂ solution is 0.67 M

4 0

4 years ago

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