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nasty-shy [4]
2 years ago
9

Amino acids are the building blocks of proteins. The simplest amino acid is glycine (H2NCH2COOH). Draw a Lewis structure for gly

cine. (Hint: The central atoms in the skeletal structure are nitrogen bonded to carbon, which is bonded to another carbon. The two oxygen atoms are bonded directly to the right-most carbon atom.) Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all hydrogen atoms and nonbonding electrons.
Chemistry
1 answer:
VashaNatasha [74]2 years ago
3 0

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

         H

          |

H₂N - C - C =O

          |      \

         H      OH

The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

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A sample of bismuth has a mass of 343g and volume of 35.0 cm3. what is the density of bismuth
elena-14-01-66 [18.8K]
I don’t think I could answer this sorry.........
4 0
2 years ago
The bond dissociation energy to break 4 bond(s) in 1 mole of CH₄ molecules is:_____ **Any help would be greatly appreciated!**
frozen [14]

Answer:

The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ

Explanation:

Since there are 4 C-H bonds in CH₄, the bond dissociation energy of 1 mol of CH₄ is 4 × bond dissociation energy of one C-H bond.

From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ

So, the bond dissociation energy to break 4 bonds in 1 mol of CH₄ is 1644 kJ

5 0
2 years ago
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with
stepan [7]

Answer:

1) Net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2) 0.765 M is  the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq) ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq) ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq) ..[3]

H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)

From one:[1] ,[2] and [3]:

2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)

Cancelling common ions on both sides to get net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2CO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL

Putting values in above equation, we get:

M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M

0.765 M is  the molarity of the carbonic acid solution.

6 0
2 years ago
One mole of h2s gas escapes from a container by effusion in 77 seconds how long would it take one mole of nh3 gas to escape from
gizmo_the_mogwai [7]
The molar masses of H2S and NH3 are 34 and 17 g/mol, respectively. The equation that would best represent the given is,
                                  Rate A/Rate B = √(molar mass B/molar mass A)
Substituting,
                               x/77 = √(17 /34 )
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Thus, it will take 54.4 seconds for NH3 to travel through the container. 
3 0
2 years ago
Read 2 more answers
Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.
Dimas [21]

Answer:

The concentration of the solution, C=7.2992\times 10^{-6} M

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

A=\varepsilon Cl

Where,  

A is the absorbance  of the solution

ɛ is the molar absorption coefficient (L.mol^{-1}.cm^{-1})

C is the concentration (mol^{-1}.L^{-1})

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 M^{-1}.cm^{-1}

l = 1 cm

Applying in the above formula for the calculation of concentration as:

A=\varepsilon Cl

0.20= 27400\times C\times 1

C = \frac{0.20}{27400\times 1} M

So , concentration is:

C=7.2992\times 10^{-6} M

4 0
2 years ago
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