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Licemer1 [7]
2 years ago
5

The organic compounds that are divided into two types, simple and complex, are called _____.

Chemistry
1 answer:
Sati [7]2 years ago
6 0

The organic compounds that are divided into two types, simple and complex, are called carbohydrates.

Carbohydrates are diveded into twy types: simple and complex (starches, fiber, glycogen).

Simple carbohydrates are made of one (monosaccaharides) or two sugar units.

Complex carbohydrates are made up of many sugar units.

For example, glucose (C₆H₁₂O₆) is a simple carbohydrate.

Glucose is chemical compound composed of six carbon atoms, twelve hydrogen atoms and six oxygen atoms.

Starch is a polymeric carbohydrate consisting of a large number of glucose units bonded by glycosidic bond. Starch is a white, tasteless and odorless powder that is insoluble in cold water or alcohol.

More about carbohydrates: brainly.com/question/20290845

#SPJ4

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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
What volume of air at 25°C and 1.00 atm can he stored in a 10.0 L high-pressure air tank if compressed to 25°C and 175 atm?
DaniilM [7]

Answer:

1750L

Explanation:

Given

Initial Temperature = 25°C

Initial Pressure = 175 atm

Initial Volume = 10.0L

Final Temperature = 25°C

Final Pressure = 1 atm

Final Volume = ?

This question is an illustration of ideal gas law.

From the given parameters, the initial temperature and final temperature are the same; this implies that the system has a constant temperature.

As such, we'll make use of Boyle's Law to solve this;

Boyle's Law States that:

P₁V₁ = P₂V₂

Where P₁ and P₂ represent Initial and Final Pressure, respectively

While V₁ and V₂ represent Initial and final volume

The equation becomes

175 atm * 10L = 1 atm * V₂

1750 atm L = 1 atm * V₂

1750 L = V₂

Hence, the final volume that can be stored is 1750L

5 0
2 years ago
What is the type of weak bond between the hydrogen of one molecule and the nitrogen of another molecule, where the two don't act
Aliun [14]
<h2>Hydrogen Bonding</h2>

Explanation:

  • Hydrogen bonding is a type of weak bond.
  • Hydrogen Bonding occurs when the hydrogen atom of one molecule is bonded with a molecule of high electronegativity element like nitrogen of another molecule.
  • It is a type of dipole-dipole interations.
  • The nitrogen is a highly electronegative element that forms a weak bond with the hydrogen atom of another molecule with dipole-dipole interaction.
  • Hydrogen bonds are weaker than covalent and ionic bonds.
  • Therefore the type of bond is Hydrogen bond.
8 0
3 years ago
If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
How does an atom change if all of its electrons are removed?<br> .
Soloha48 [4]

Answer:

If an atom looses all of its electrons then it will become positively charged. It will also turn into an Ion.

Explanation:

4 0
3 years ago
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