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Damm [24]
3 years ago
5

Describe the process of making the four major macromolecules.

Chemistry
2 answers:
dybincka [34]3 years ago
8 0

Answer:

<em>Proteins, carbohydrates, nucleic acids, and lipids are the four major classes of biological macromolecules—large molecules necessary for life that are built from smaller organic molecules. Macromolecules are made up of single units known as monomers that are joined by covalent bonds to form larger polymers.</em>

Explanation:

Have a great day!

umka2103 [35]3 years ago
8 0

Answer:, , nucleic acids, and lipids are the four major classes of biological macromolecules—large molecules necessary for life that are built from smaller organic molecules. Macromolecules are made up of single units known as monomers that are joined by covalent bonds to form larger polymers. I forgot to add carbohydrates and proteins also my bad

i just started learning aboit this too

Explanation:

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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
1. Compare and contrast the characteristics of metals and nonmetals.
Alona [7]

Answer:

1.Metals

These are very hard except sodium

These are malleable and ductile  pieces

These are shiny

Electropositive in nature

Non-metals

These are soft except diamond

These are brittle and can break down into pieces

These are non-lustrous except iodine

Electronegative in nature

2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.

3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).

4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.

4 0
3 years ago
Read 2 more answers
Substance X has a fixed volume, and the attraction between its particles is strong. Substance Y has widely spread out particles
ycow [4]

Answer: Option (d) is the correct answer.

Explanation:

In solids, molecules are held together by strong intermolecular forces of attraction. As a result, they are unable to move from their initial position and can only vibrate at their mean position.

Hence, a solid has definite shape and volume. Solids cannot be compressed.

Whereas in plasma, molecules are hot ionized which include positively charged ions and negatively charged electrons. They collide much more rapidly with each other and are widely spreaded out.

Therefore, they occupy the volume of container in which they are placed. Plasma can be compressed.

Thus, we can conclude that substance X is a solid and substance Y is a plasma.

6 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
How many grams of AgNO3 are needed to prepare a 0.25 m solution in 500 grams of water?
tekilochka [14]
21.25 grams would be the answer. I agree
7 0
3 years ago
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