CHEMISTRY HIGH SCHOOL

Which of the following is an oxidation-reduction reaction? Fe2O3 + 3CO mc009-1.jpg 2Fe + 3CO2 CuSO4 + 2NaOH mc009-2.jpg Cu(OH)2 + Na2SO4 2NaOH + H2CO3 mc009-3.jpg Na2CO3 + 2NaOH Pb(NO3)2 + Na2SO4 mc009-4.jpg 2NaNO3 + PbSO4

Answers

Answer 1

Answer:

Answer:

$Fe_(2)O_(3)+3CO\rightarrow 2Fe+3CO_(2)$ is an oxidation-reduction reaction.

Explanation:

In an oxidation-reduction reaction, the oxidation number of some elements must change during reaction.

$Fe_(2)O_(3)+3CO\rightarrow 2Fe+3CO_(2)$ represents an oxidation-reduction reaction.

oxidation number

Fe in $Fe_(2)O_(3)$ +3

Fe in Fe 0

C in CO +2

C in $CO_(2)$ +4

SO, oxidation number of Fe decreases during reaction. Hence $Fe_(2)O_(3)$ is being reduced. Also, oxidation number of C increases during reaction. Hence CO is being oxidized.

In all other given reactions, oxidation number of elements are not changed during reaction. Hence they are not oxidation-reduction reaction.

Answer 2

Answer:

Answer: it's A

Explanation:

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The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 3.00 g of butane?

Answers

First, calculate the number of mols of C4H10 in 3.00 g of C4H10.

3.00g C4H10 / 58.1 g C4H10 = .0516 mol C4H10
There are 4 atoms of carbon per mol of C4H10
Multiply .0516 mol C4H10 by 4 (number of atoms of carbon)=.20654 mol Carbon.
To convert to atoms, multiply that by avogadro's number, 6.022 x 10^23
which equals 1.2438 x 10^23 atoms of Carbon

Answer:

1.2433 *10^23 carbon atoms

Explanation:

First you need to calculate the grams of the empirical formula.

C4H10

Carbon = 12.01 g x 4

Hydrogen = 1.008 x 10

Sum those together to get = 58.12 g

Divide the 3.00g by the total g in the compound.

3.00/58.12 = 0.0516

Then times 0.0516 by the number of carbon atoms

0.0516 x 4 = .20646