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hichkok12 [17]
3 years ago
12

Draw the Lewis structure of NCl3NCl3 . Include lone pairs. Select Draw Rings More Erase Select Draw Rings More Erase Select Draw

Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Cl N
Chemistry
1 answer:
Feliz [49]3 years ago
6 0

Answer:

See explanation and image attached

Explanation:

Nitrogen trichloride NCl3 contains one nitrogen and three chlorine atoms. Each chlorine atom has three lone pairs of electrons while nitrogen has one lone pair of electrons.

The molecule geometry of the molecule is trigonal pyramidal while it's electron domain geometry is tetrahedral. Nitrogen is the central atom and is found in sp3 hybridization.

There is no formal charge on the molecule.

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Elements Y and elements Z would have similar properties due to the fact that they both posses the same number of valence electrons. They both have a single valence electron that determines the corresponding elements bonding properties and the fact that it can either donate 1 valence electron to produce an ion that would be attracted to another atom, that is also an ion. Assuming that these elements are group 1 elements, they do not undergo in covalent bonding.
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Which of these elements is this group?
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Answer:

Strontium

Explanation:

In the periodic table, an element with two (2) valence electrons is found on group 2. Group 2 is a group of the periodic table that harbors element called ALKALINE EARTH METALS. As the name implies, they are metals that possess shiny and solid characteristics at room temperature.

Group 2 elements include the following: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), barium (Ba), and radium (Ra). Based on the descriptive information in this question, the element being described is a GROUP 2 element. Based on the elements in the option, only STRONTIUM (Sr) is a group 2 element.

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3 years ago
Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Cu^{2+}/Cu]}=+0.34V

E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

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3 years ago
Which of the following is NOT a property of an acid?
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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
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