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Elena L [17]
4 years ago
10

How many pairs of electrons are shared between the nitrogen atoms in a molecule of n2?

Chemistry
1 answer:
Kobotan [32]4 years ago
5 0
An N∨2 Molecule shares 3 pairs of electrons and forms a triple bond. Each N atom contributes 3 electrons making three pairs of electrons.
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Which element is placed in the same period as ruthenium but has a higher atomic number than it?
emmasim [6.3K]

Answer:

The part that is put in the same time as ruthenium, but has a greater number of atoms than silver.

8 0
3 years ago
Calculate the solubility (in m) of cobalt(ii) hydroxide, co(oh)2(s) in h2o. ksp = 1.00×10-16 at a specific temperature.
antoniya [11.8K]
according to this reaction:

by using the ICE table:

              CO(OH)2 → CO2+  +  2OH-

intial                                0             0

change                           +X           +2X

Equ                                   X              2X

when Ksp = [CO2][OH-]^2

by substitution:

3.6 x 10^-16 = X * (2X)^2

3.6 x 10^-16 = 4 X^3

∴X = 3.88 x 10^-6 M
8 0
4 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

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

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

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

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

6 0
3 years ago
Read 2 more answers
Considering the limiting reactant, what is the mass of zinc sulfide produced from 0.250 g of zinc and 0.750 g of sulfur? Zn(s)+S
DIA [1.3K]

Answer:

The mass of zinc sulfide produced is  M_{ZnS} =  0.76 \ g

Explanation:

From the question we are told that

   The mass of zinc is  m_z =  0.750 \ g

    The mass of sulfur is  m_s =  0.250 \ g

The molar mass   of  Zn_{(s)}  is a constant with value  65.39 g /mol

The molar mass of S_{(s)}  is a constant with value  32.01 g/mol

The molar mass of  ZnS_{(s)} is a constant with value 97.46  g/mol

The reaction is  

        Zn_{(s)} + S_{(s)}  ------> ZnS_{(s)}

   So from the reaction

       1 mole of  Zn_{(s)} react with 1 mole of  S_{(s)} to produce 1 mole of ZnS_{(s)}

This implies that

65.39 g /mol of  Zn_{(s)} react with 32.01 g/mol of  S_{(s)} to produce   97.46  g/mol  of ZnS_{(s)}

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  Zn_{(s)} react with 0.250 of  S_{(s)} to produce x \  g of  ZnS_{(s)}

So

      x =  \frac{97.46 * 0.250}{32.01}

       x =  0.76 \ g

Thus the mass of the mass of zinc sulfide produced is

    M_{ZnS} =  0.76 \ g

 

     

7 0
4 years ago
Which two Statements explain how a cells parts help it get nutrients.
Shalnov [3]

Answer:

The mitochondria change energy in organic compounds into a

form the cell can use.

The chloroplasts take in energy from sunlight and change it into

organic matter.

Explanation:

8 0
3 years ago
Read 2 more answers
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