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Rasek [7]
3 years ago
5

Consider the reaction.

Chemistry
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

2.1 x 10^-2

Explanation:

Divide 8.4 x 10^-3 x 8.4 x 10^-3 by 5.82 x 10^-2 and you end up with 2.1 x 10^-2

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A certain first order reaction has a half-life of 54. 3 s. How long will it take (in s) for the reactant concentration to decrea
Maksim231197 [3]

Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

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A ________ reaction occurs when one compound reacts and is broken down into different elements or simpler compounds?
faust18 [17]

Answer: decomposition

Explanation:

1 . Combustion is a type of chemical reaction in which a hydrocarbon reacts with oxygen to form carbon dioxide ans water along with liberation of large amount of energy.

2. Decomposition is a type of chemical reaction in which a single reactant gives two or more than two products.

CaCO_3\rightarrow CaO+CO_2

3. Single replacement is a type of chemical reaction in which a more reactive element displaces the less reactive element from its slat solution.

4. Synthesis is a type of chemical reaction in which two or more than two reactants combine together to give a single product.

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3 years ago
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Describe the difference between pure substance and mixtures ( give an example each)
kondaur [170]

Answer:

a pure substance consists only of one element or one compound

a mixture consists of two or more different substances, not chemically joined together.

examples:

pure substance : Hydrogen gas - Diamond - Gold metal.

mixture : water and oil - mixtures of sand and water - trail mix

Explanation:

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2 years ago
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Color is not very helpful in mineral identification because _____.
zhannawk [14.2K]
Some minerals tend to look alike.
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3 years ago
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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 Zn
nlexa [21]

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}-\frac{0.0592}{n}\log \frac{[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-\frac{0.0592}{2}\log \frac{(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}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

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

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(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}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

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

0.45=0.53-\frac{0.0592}{2}\log \frac{(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

5 0
3 years ago
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