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Schach [20]
3 years ago
9

What is the mass in grams of 13.2 amu?

Chemistry
1 answer:
topjm [15]3 years ago
3 0

Amu or also known as atomic mass unit is a SI unit that is used in chemistry to identify the total of the mass of neutrons and protons. In this case,a carbon-12 has a mass of amu. amu is equivalent to grams per mole. In this case, the mass of 13.2 amu is 13.2 grams per mole
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For each of the bonds listed, tell which atom is more negatively charged a. C--H b. C--N c. C--S d. C--O
blondinia [14]

Explanation:

The atom is considered to be more electronegative , when it has the tendency to get a slight negative i.e. , delta negative charge on it , and the other atom gets a delta positive charge .

In the periodic table ,

As we go left to right in the periodic table , the electronegativity increases ,

and

As we go top to bottom , the electronegativity decreases .

Hence , from the option given in the question ,

a. C--H , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , Carbon is more electronegative than hydrogen .

b. C--N ,  the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , nitrogen is more electronegative than Carbon .

c. C--S , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , sulfur is more electronegative than Carbon .

d. C--O , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , oxygen is more electronegative than Carbon .

5 0
3 years ago
In the equilibrium and concentration gizmo, you learned that you can predict the direction of a reaction by comparing the reacti
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And the significant amount of volume can be differed by its solitude

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3 years ago
hydrolysis of decapeptide P with the enzyme trypsin affords the following fragments: Glu-Gly-Lys, Gln-Val-Ile, Ala-Ser-Phe-Lys.
ehidna [41]

Answer:

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

Explanation:

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named trypsin:

  • Glu-Gly-Lys,
  • Gln-Val-Ile
  • Ala-Ser-Phe-Lys

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named chymotrypsin:

  • Lys-Gln-Val-Ile,
  • Glu-Gly-Lys-Ala-Ser-Phe

In order to determine the sequence of protein P , we will arrange fragments in such a way so that common fragments or the common parts of fragments should come under each other.

On arranging these fragments :

Glu-Gly-Lys-Ala-Ser-Phe

Glu-Gly-Lys

                   Ala-Ser-Phe-Lys

                                         Lys-Gln-Val-Ile

                                                Gln-Val-Ile

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

3 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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