CHEMISTRY HIGH SCHOOL

Name two metals that do not always lose the same number of electrons

Answers

Answer 1
Answer:

Copper and Iron are two metals that can lose different numbers when they react with another molecule.

What are transition metals?

  • These are the metals that have variable valence electrons and form many-colored co-ordinate compounds.

  • The transition metals have variable oxidation state means they can lose different numbers of electrons while making a bond.

  • The variable oxidation state is due to the electronic configuration in which they have partially filled d- block which allows them to have variable oxidation numbers.

Therefore, copper and Iron are two metals that can lose different numbers when they react with another molecule.

Learn more about Transition metals:

brainly.com/question/8098159
Answer 2
Answer: it would be copper and iron -A

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Scientists test a hypothesis by?
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If a solution is made by mixing 50.0 moles of KOH to 3.4 kilograms of water then what is the solutions molality?

.A pure gas that is 14.3% hydrogen and 85,7% carbon by mass has a density of 2.5 g L-' at 0°C and 1 atm
pressure. What is the molecular formula of the gas

Answers

Answer:

C

Explanation:

4× CH2 = C4H8

Any substance that yields a hydrogen ion when placed in a water solution is called a(n) _____.acid
base
hydronium ion
hydroxide ion

Answers

Answer : Option A) Acid


Explanation : Any substance that yields a hydrogen ion when placed in a water solution is called an acid.


According to Arrhenius theory of Acids and Bases, Acids are those substances which on dissociation in solution generates H^(+) ions. Whereas a Base is a substance that dissociates in the solution to produce OH^(-) ions.

Answer:

the answer is acid.

Explanation:

Discuss the shapes of Nh3 and H2O on the basis of VESPER theory?​

Answers

Answer:

Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

Explanation:

Which of the following is an example of radiation?A. boiling water on the stove
B. sun bathing outside on a sunny day
C. touching a stove and getting burned
D. warm and cool water currents in the ocean

Answers

I believe the best answer to that question wud be D. I cud b wrong 

A similarity between the forces involved in both ionic and covalent bondingis that bothO A) involve the sharing of electrons.O B) involve the transfer of electrons.O C) involve the attraction of opposite charges.requre same amountof energy toereakther bonds

Answers

Stating the difference between covalent and ionic bonding....

ionic bonds consist of the transfer of electrons, and covalent consists of the sharing of electrons.

This being said..answer choices A and B are out of the picture.

Thus, answer choice C is your answer.

Involve the attraction of opposite charges;require same amount of energy to break their bonds.

77. 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 Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

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)-(0.0592)/(n)\log ([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-(0.0592)/(2)\log ((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)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])

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

E_(cell)=0.53-(0.0592)/(2)\log ((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)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])

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

0.45=0.53-(0.0592)/(2)\log ((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
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