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Gnom [1K]
3 years ago
6

Why the ph of glycine increases when 0.1 M NaOH is added dropwise​

Chemistry
1 answer:
shtirl [24]3 years ago
7 0

Answer:

The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

Explanation:

The glycine is an amino acid with the following chemical formula:

NH₂CH₂COOH  

The COOH functional group is what gives the acid properties in the molecule.      

Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:

NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O

The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.  

Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.  

I hope it helps you!  

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The atomic masses of 35-Cl (75.53 percent) and 37-Cl (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the
AlladinOne [14]

Answer:

35.45 amu is the average atomic mass of chlorine.

Explanation:

The average atomic mass is given :

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Atomic mass of 35-Cl = 34.968 amu

Percentage abundance of 35-Cl = 75.53%

Fractional abundance of 35-Cl = 0.7553

Atomic mass of 37-Cl = 36.956 amu

Percentage abundance of 37-Cl = 24.47%

Fractional abundance of 37-Cl = 0.2447

Average atomic mass of chlorine:

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3 years ago
given the following chemical equation, determine how many grams of N2 are produced by 9.24 of H2O2 nd 6.56g of N2H4
taurus [48]
I think the chemical reaction is:<span>

N2H4 + 2 H2O2-> N2 + 4H2O

We are given the amount reactants allowed to react. This will be the starting point of the reaction. First, is to find the limiting reactant.

9.24 g H2O2 ( 1 mol / 34.02 g ) = 0.27 mol H2O2
6.56  g N2H4 ( 1mol / 32.06) = 0.20 mol N2H4

Since from the reaction we have 1:2 ratio of the reactants then the limiting reactant is hydrogen peroxide. We will use this to find the amount of N2 produced.

0.27 mol H2O2 ( 1 mol N2 / 2 mol H2O2 ) ( 14.01 g N2 / 1 mol N2) =1.89 g N2 </span>
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