They can be joined through a Peptide bond
The amino acids serve as the foundation for proteins. Water is produced when the amino acids are linked to form a lengthy chain of acids via amino and carboxyl. The main protein is made up of these long chain amino acids.
When the carboxyl group of one molecule combines with the amino group of the other molecule, a molecule of water is released, and a peptide bond is created between the two molecules (H2O). This condensation event, sometimes referred to as a dehydration synthesis reaction, typically takes place between amino acids.
<h3>What is a Peptide bond ?</h3>
The carboxyl group of one amino acid is joined to the amino group of another to produce a peptide bond, also known as a eupeptide bond. In essence, a peptide link is an amide-type covalent chemical bond.
Learn more about Peptide bond here:
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Answer:
![V_1=23.3~mL](https://tex.z-dn.net/?f=V_1%3D23.3~mL)
Explanation:
In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:
![C_1*V_1=C_2*V_2](https://tex.z-dn.net/?f=%20C_1%2AV_1%3DC_2%2AV_2)
Now we can identify the variables:
![C_1=~1.475_M](https://tex.z-dn.net/?f=C_1%3D~1.475_M)
![V_1=~?](https://tex.z-dn.net/?f=V_1%3D~%3F)
![C_2=~0.1374_M](https://tex.z-dn.net/?f=C_2%3D~0.1374_M)
![V_2=~250.0~mL](https://tex.z-dn.net/?f=V_2%3D~250.0~mL)
If we plug all the values into the equation:
![1.475_M*V_1=0.1374_M*250.0~mL](https://tex.z-dn.net/?f=1.475_M%2AV_1%3D0.1374_M%2A250.0~mL)
And we solve for
:
![V_1=\frac{0.1374_M*250.0~mL}{1.475_M}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7B0.1374_M%2A250.0~mL%7D%7B1.475_M%7D)
![V_1=23.3~mL](https://tex.z-dn.net/?f=V_1%3D23.3~mL)
I hope it helps!
The density of the sample is:
Density = mass / volume
Density = 9.85 / 0.675
Density = 14.6 g/cm³
If the sample has 95% gold, and 5% silver, its density should be:
0.95 x 19.3 + 0.05 x 10.5
Theoretical density = 18.9 g/cm³
The difference in theoretical and actual densities is very large, making it likely that the jeweler was not telling the truth.
Answer:
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Explanation:
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5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.
<u>Explanation</u>:
- Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:
Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃
- From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of H₂SO₄.
- Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
- To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is = 5.451 X 10³ kg.