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d1i1m1o1n [39]
3 years ago
14

A compound contains 57.2 percent carbon, 6.1 percent hydrogen, 9.5 percent nitrogen, and 27.2 percent oxygen. What the empirical

formula of the compound?
Chemistry
1 answer:
Julli [10]3 years ago
4 0

Answer:

So the empirical formula is C14H18N2O5

Explanation:

C = 57.2% = 12g/mol

H = 6.1% = 1g/mol

N = 9.5% = 14g/mol

O = 27.2% = 16g/mol

Empirical Formula for compound hmm

Assume

C = 57.2g

H = 6.1g

N = 9.5g

O = 27.2g

So we have

C = 57.2g/12g =  4.76  moles

H = 6.1g/1g      =  6.10   moles

N = 9.5g/14g   =  0.68  moles

O = 27.2g/16g =  1.70   moles

Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number.

C =  4.76  moles / 0.68 moles = 7

H = 6.10   moles / 0.68 moles = 9

N = 0.68  moles / 0.68 moles = 1

O = 1.70   moles / 0.68 moles = 2.5

Ok so we now have the ratios but for O it's 2.5, have to be whole numbers so we will need to double all the numbers.

C = 14

H = 18

N =  2

O =  5

So the empirical formula is C14H18N2O5

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7 0
2 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

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

The half-cell reactions are:

Oxidation half reaction (anode):  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Pb^{2+}+2e^-\rightarrow Pb

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

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

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

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Therefore, the cell potential for this reaction is 0.50 V

5 0
4 years ago
Read the information about the halogen family.
Elodia [21]

Answer:

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Explanation:

Chlorine is a member of the halogen family known as a toxic yellowish green gas. Inhalation of chlorine for a prolonged period of time leads to  pulmonary edema. If a person comes in contact with compressed liquid chlorine the person may experience frostbite of the skin and eyes.

However chlorine is very useful in water disinfection and is preferred in water treatment because it provides residual disinfection of the treated water.

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IceJOKER [234]

Answer:

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