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QveST [7]
3 years ago
7

Identify the oxidizing and reducing agents in this reaction: Cr2O3(s) + 2Al(s) 2Cr(s) + Al2O3(s)

Chemistry
2 answers:
DiKsa [7]3 years ago
7 0

In this reaction, Cr2O3 is the oxidizing agent which will oxidize Al to Al2O3.

Al is the reducing agent that will reduce Cr2O3 to Cr.

An oxidizing agent has the property of oxidizing another molecule (giving it one or more oxygen atoms) by being reduced.

A reducing agent has the property of reducing another molecule (by removing one or more oxygen molecules) by being oxidized.

aivan3 [116]3 years ago
3 0
Cr2O3+ 2Al -----------> 2Cr+ Al2O3

Cr2O3--------2Cr.
( oxidation no. of cr +3 to +1 means hain of electron so it act as oxidizing agent)

similarly,

2Al------> Al2O3
( oxidation no. of Al +1 to +3 means loss of electron so it act as reducing agent)
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Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
Genrish500 [490]

Explanation:

The given data is:

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The concentration of reactant after 8 hrs can be calculated as shown below:

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k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

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Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0
3 years ago
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Diano4ka-milaya [45]
The balanced chemical reaction is given as follows:

<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
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3 years ago
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