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2 years ago

Amines are ________. brønsted-lowry bases brønsted-lowry acids neutral in water solution unreactive

Chemistry

1 answer:

sammy [17]2 years ago

8 0

Answer:

A.) Brønsted-Lowry bases

Explanation:

Amines have a lone pair of electrons.

Brønsted-Lowry bases donate a lone pair of electrons in exchange for a hydrogen ion.

Therefore, if exposed to an acid, amines will give up electrons in order to bond with a hydrogen. This makes them Brønsted-Lowry bases.

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What is the "magic number' of valence electrons? How does this influence the type of bonds formed by various combinations of

svp [43]

Magic number is any number in electron shells that suggest stability. It corresponds to total number of electrons in filled electron shells.

If an electron is having magic number, then it forms stable bonds.

Explanation:

The magic numbers are 2,10,18,36,54, 86 and 126. This refers to the total number of electrons that an electron can have when it is completely filled.

Atomic nuclei which carries either of these nucleons have high binding energy as compared to others. Hence, they have high stability. Bonds in such elements are more strong.

Radioactive decay of such elements is very slow.

Eugene Winger coined the term "magic number".

6 0

3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

4 years ago

Which kind of bond is formed when two atoms share electrons to form a molecule?

Zina [86]

A covalent bond is formed.

6 0

4 years ago

A ball is moving at a speed of 6.70 m/s. If the kinetic energy of the ball is 3.10 J, what is the mass of the ball?

Anni [7]

Answer:

Explanation:

The mass of the ball can be found by using the formula

$m = \frac{2k}{ {v}^{2} } \\$

v is the velocity

k is the kinetic energy

From the question we have

$m = \frac{2(3.10)}{ {6.70}^{2} } = \frac{6.20}{44.89} \\ = 0.138115...$

We have the final answer as

Hope this helps you

7 0

3 years ago

the roof of a building is 0.2km^2. during a rainstorm, 5.5 cm of rain was measured to be sitting on the roof. what is the mass i

Nikolay [14]

The mass in kg of the water on the roof after the rainstorm is mathematically given as

The mass of the water on the roof after a rainstorm is 1.1 *10^{10}g or 1.1*10^{7}kg

<h3>What is the mass in kg of the water on the roof after the rainstorm? </h3>

Generally, the equation for the Area of the roof is mathematically given as

$A=0.20 \times 10^{10} \mathrm{~cm}^{2}$

height of the rain =5.5cm

the volume of the rain on the roof =1.1 * 10^{10} CC

Generally, the equation for mass is mathematically given as

mass=volume*density

$&=\left(1.1 \times 10^{10} \mathrm{cC} \times 1 \mathrm{~g} / \mathrm{cc}\right) \\&=1.1 \times 10^{10} \mathrm{~g}$

In conclusion, the Mass of the water on the roof after a rainstorm is 1.1 *10^{10} or $1.1*10^{7}