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soldier1979 [14.2K]
3 years ago
15

How many atoms are in 4 Al2(SO3)3

Chemistry
1 answer:
scZoUnD [109]3 years ago
4 0

There are 337.23 × 10²³ atoms in 4 moles of aluminum sulfite Al₂(SO₃)₃.

Explanation:

The questions ask how many atoms are in 4 moles of aluminum sulfite Al₂(SO₃)₃?

To answer this we use the Avogadro's number to devise the following reasoning:

if in        1 mole of Al₂(SO₃)₃ there are 14 × 6.022 × 10²³ atoms

then in  4 moles of Al₂(SO₃)₃ there are X  atoms

X = (4 × 14 × 6.022 × 10²³) / 1 = 337.23 × 10²³ atoms

Learn more about:

Avogadro's number

brainly.com/question/13848085

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#learnwithBrainly

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Answer:

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3 years ago
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2 years ago
The relative strengths of covalent bonds and van der Waals interactions remain the same when tested in a vacuum or in water. How
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Answer and Explanation:

The explanation given in the problem is correct but not totally encompassing.

Van der waals interactions are a type of hydrophobic interaction, in which they do not interact with the polar water molecule. Covalent bonds involve the sharing of electrons between atoms of relatively similar electronegativities, and are most often too strong to disrupt by polar molecules of water. Therefore, covalent bonds and van der waals forces have an Intrinsic bond strength value that is independent of the environment.

However, either the partial negative oxygen atom or the partial positive hydrogen atoms in water molecules disrupt hydrogen or ionic bonds. Water is known to form hydrogen bonds with other polar or charged molecules, thus reducing the strength of interaction these molecules would normally have in the absence of water. Basically, these compounds with Hydrogen or Ionic bonds ionize, whether partially or fully in water, thereby leading to a decrease in bond strength in water.

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7 0
3 years ago
What is the best way to<br> determine which brand of<br> paper towel is the<br> strongest when wet?
luda_lava [24]

Answer:Bounty

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4 0
3 years ago
The radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-g sample of radon- (3 222 would be left after approxi
Novay_Z [31]

Answer:

1.115 g

Explanation:

Applying,

R = R'2^{n/t}................... Equation 1

Where R = original sample of radon-222, R' = sample of radon-222 left after decay, n = Total time, t = half-life.

make R' the subject of the equation

R' = R/(2^{n/t})............... Equation 2

From the question,

Given: R = 73.9 g, n = 23 days, t = 3.8 days.

Substitute these values into equation 2

R' = 73.9/(2^{23/3.8})

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R' = 1.115 g

6 0
3 years ago
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