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Firdavs [7]
2 years ago
8

What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed as

Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
5 0

Answer:

13.5

Explanation:

Molar mass of Fe2O3 = 2*56 + 16*3 = 112 + 48 =160 approximate value

Number of moles of Fe2O3 = mass/molar mass = 40/160 = 0.25 mol

The reaction is balanced. Fine

1 mol of Fe2O3--------> 2 mol of Al

0.25 mol--------> 0.25*2 = 0.5 mol of Al

Mass of Al required = Number of moles * Molar mass

= 0.5 * 27 = 13.5 g

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In the sn2 experiment, what was the purpose of washing the distilled product with 5% naoh
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We must __________ forces to calculate net force if the forces are going in OPPOSITE directions.
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Answer:

subtract

Explanation:

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5 0
2 years ago
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

4 0
3 years ago
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