Atoms in the amino acids become the h₂O molecule produced by their action in the model and come off from the central carbon and nitrogen but not from the carboxyl, R side chain, or amine.
An amino acid is a group of organic molecules that consist of a basic acidic carboxyl group (―COOH), amino group (―NH2), and an organic R group (or side chain) that is different from each amino acid. Amino acid, the term is a short form of α-amino [alpha-amino] carboxylic acid.
Whereas, the peptide bond is the chemical bond which is a chemical bond formed between two molecules when the carboxyl group of a particular molecule reacts with the amino group of the other molecule, leading to releasing a molecule of water (H2O).
Each molecule consists of a central carbon atom referred to as the α-carbon, to which both a carboxyl group and amino are attached. The remaining two bonds of the α-carbon atom are generally occupied by the R group and a hydrogen (H) atom .
To know more about amino acids refer to the link brainly.com/question/14583479?referrer=searchResults.
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The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,
Mg + 2HCl --> H2 + MgCl2
The number of moles of HCl that is needed for the reaction is calculated below.
n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
n = 0.0385 mols HCl
From the given concentration, we calculate for the required volume.
V = 0.0385 mols HCl/(0.650 mols/L)
V = 0.05926 L or 59.26 mL
<em>Answer: 59.26 mL of HCl</em>
Answer:
Explanation: Sodium Nitroprusside, whose molecular formula is • 2H2O, and whose molecular weight is 297.95. Dry sodium nitroprusside is a reddish-brown powder, soluble in water.
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
Best regards.
Answer:
pH = 7.233
Explanation:
Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.
NaHSO₃ reacts with NaOH thus:
NaHSO₃ + NaOH → Na₂SO₃ + H₂O
50.0 mL of 1.00 M NaOH are:
0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced 0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:
NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>
Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>
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As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:
pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]
pH = 7.167 + log₁₀ [0.184] / [0.158]
<em>pH = 7.233</em>
