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3 years ago

A zinc salt, E when

Chemistry

1 answer:

Naily [24]3 years ago

6 0

Answer:

The salt is Zn(NO3)2

When the residue is allowed to cool, its colour changes from yellow to white

The equation of the reaction is; 2Zn(NO3)2(s) --->2ZnO(s) + 4NO2(g) + O2(g)

Explanation:

Decomposition of Zn(NO3)2 is one of the routes by which ZnO is obtained. ZnO is a white insoluble powder which has a yellow colour when hot and a white colour when cold. It is an amphoteric substance used as a white pigment in paints and as a filler for rubber.

In a very pure form, it is used in cosmetic powders and creams as well as in medicinal ointments and lotions.

Also recall that NO2 is a brown gas and oxygen gas rekindles a glowing splint.

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5. Complete and balance an equation for each reaction:

Kipish [7]

CaI₂ + Hg(NO₃)₂ --------->HgI₂ + Ca(NO3)2

2Al + 3Cl₂ --------->2AlCl3

Ag + HCl ------->AgCl + H2

C2H2 + 5O2 --------> 4CO2 + 2H2O

MgCl₂ --------->Mg + Cl2

6 0

3 years ago

5/3 * 2/3 what's the answer

Contact [7]

5 x 2 = 10

3 x 3 = 9

10/9, or 1 1/9 is your answer

hope this helps

6 0

3 years ago

The nucleus of an atom is made up of a dense ball

Marina86 [1]

Answer:

The nucleus of an atom is made up of a dense ball

of protons and neutrons surrounded by electrons.

Explanation:

We know that an atom consist of electrons, protons and neutrons. Neutrons and protons are present inside the nucleus while electrons are present out side the nucleus. Electron has a negative charge and is written as e⁻. The mass of electron is 9.10938356×10⁻³¹ Kg . While mass of proton and neutron is 1.672623×10⁻²⁷Kg and 1.674929×10⁻²⁷ Kg respectively.

Symbol of proton= P⁺

Symbol of neutron= n⁰

The number of electron or number of protons are called atomic number while mass number of an atom is sum of protons and neutrons. The umber of protons and electrons are always equal to make the atom electrically neutral and when an atom loses its valance electron the number of protons increases and thus positive charge increased and atom form cation.

7 0

3 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

olga_2 [115]

Answer: The $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

Explanation:

For the given half reactions:

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V$

Net reaction: $Fe(s)+Cu^{2+}\rightarrow Fe^{2+}+Cu(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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

Putting values in above equation, we get:

$E^o_{cell}=0.34-(-0.44)=0.78V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.78 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}$

Hence, the $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

3 0

4 years ago

Please help me with this activity!! I have to submit it ASAP. It's attached as a pdf. Not too long and for an Honors Electrons l

brilliants [131]

(1) Seven electrons

1s² 2s²2p³

There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.

(2) 22 electrons

1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²

There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.

(3) 17 electrons

1s² 2s²2p⁶ 3s²3p⁵

There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.

(4) n = 4, l = 2

(5) N=2 I=0 MI=0

(6) There can be two electrons in one orbital maximum. The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max.

(7) n = 4 can hold up to 16 electrons

There are 4 sublevels; 42, 4p, 4d and 4f

4s (1 orbital), 4p (3 orbitals)

4d (5 orbitals), 4f (7 orbitals)

Maximum number of electrons in energy level: 16

s = 1 orbital

p = 3 orbitals

d = 5 orbitals

f = 7 orbitals

Add these up;

1 + 3 + 5 + 7 = 16

(8) 14 electrons If l = 3, the electrons are in an f subshell. The number of orbitals with a quantum number l is 2l + 1, so there are 2×3 + 1 = 7 f orbitals. Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

sorry i didn't do the last two.. also sorry if some or none of these are correct, i tried my best

7 0

3 years ago