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2 years ago

A carbon skeleton is covalently bonded to both an amino group and a carboxyl group. When placed in water it ____

Chemistry

1 answer:

kotykmax [81]2 years ago

8 0

Answer:

The correct option is: (D) would function as both an acid and a base

Explanation:

A carbon skeleton bonded to a amino group as well as a carboxyl group, will behave as an acid in basic medium and base in acidic medium. This is because the carboxyl group present in the compound will release a proton in basic medium and the amino group will accept a proton in the acidic medium.

<u>Therefore, a carbon skeleton which is covalently bonded to a carboxyl and amino group will behave as both acid and base.</u>

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

3 years ago

Al(s) + NiSO4(aq)<br> Balance

irinina [24]

Answer:

$2Al(s) + 3NiSO4 --> Al2(SO4)3 + 3Ni$

Explanation:

This reaction type is a single replacement. The format of a single replacement is:

$A+BC --> B+AC$

A= Al

B= Ni

C= SO

The coefficient 3 for Ni would become a subscript for AC. After you plug those into the reaction you need to count how many of each are on the left side and try to get the same number on the right side. Both sides must be equal to have a balanced equation.

8 0

3 years ago

What is a potential benefit of fuel cell cars?

77julia77 [94]

Answer:

no recharging necessary

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3 years ago

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James decides to walk home from school today. He lives 3 miles from school and can walk home in 45 minutes. At what rate is Jame

Natasha2012 [34]

One mile every 15 minutes