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Naily [24]
3 years ago
13

3. Which of the following sentences apply to neutrons? a. have a negative charge. b. circle around the atom nucleus. c. are the

only particles that form the nucleus. d. have a mass of 1 amu. SA 2​
Chemistry
1 answer:
DerKrebs [107]3 years ago
3 0

Answer:

d.have a mass of 1 amu..

Explanation:

Does not match a as the answer for a is electrons B.electron circle around the nucleus because neutrons are inside the nucleus.C.proton and neutrons make the nucleus so the only answer left is .d .which is correct

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Glucose is made up of hydrogen, oxygen, and carbon atoms.
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Answer:

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What is the process of discovery that allows us to link abservations with actual facts?
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Inquiry to find the answers we search for
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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
3 years ago
1. A solution is a mixture. The part of<br> the solution in the greater amount is<br> called the
olga_2 [115]

Answer:

Solvent

Explanation:

Many people use the solute to describe the solid being dissolved and the solvent to describe the thing doing the dissolving, but really solvent means the part of the solution with a greater amount/concentration.

For example, if you have 1 gram of salt in 10 liters of water, the water is the solvent.

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What is the atomic number of an atom? Why is this number important?
aalyn [17]

The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms
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