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Sav [38]
4 years ago
15

What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?

Chemistry
1 answer:
natka813 [3]4 years ago
6 0
<span>26.833 liters

Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
</span>
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List and discuss at least 4 reasons why a laboratory should operate using GLP.
Drupady [299]

Answer:

The list and discussions are stated below:

Explanation:

Good Laboratory Practice (GLP) is extremely important.

1. Organization

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2. Safety

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4 years ago
Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can
ratelena [41]

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = Length\times breadth\times height

Thus,

Volume = 304\times 0.014\times x\ mm^3=4.256x\ mm^3

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

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Which of the following is a correct observation when water is added to quick lime?
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Answer:

c

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thw observation of water that is added

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