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Snezhnost [94]
3 years ago
10

How many moles of water are in 1.23x10^18 water molecules

Chemistry
2 answers:
Olegator [25]3 years ago
8 0

Answer:

There are 2.04*10⁻⁶ moles of water in 1.23*10¹⁸ water molecules

Explanation:

Avogadro Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

In this case you know that there are 1.23*10¹⁸ molecules of there. So, knowing Avogadro's number, a rule of three applies as follows: if 6.023*10²³ molecules represent 1 mole of the compound, 1.23*10¹⁸ molecules in how many moles will they be?

moles=\frac{1.23*10^{18} molecules*1mole}{6.023*10^{23}molecules }

moles=2.04*10⁻⁶

<u><em>There are 2.04*10⁻⁶ moles of water in 1.23*10¹⁸ water molecules</em></u>

Harlamova29_29 [7]3 years ago
6 0

By 1.23 x 1024 you mean 10 to the power of 24 molecules? If so all you need to do is divide the number of molecules you have by Avagadros number, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance.


1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20


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What is the chemical formula for 8.6 mol of sulfur and 3.42 mol of phosphorus
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<h3>How do I determine the formula of the compound?</h3>

From the question given above, the following data were obatined:

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  • Phosphorus (P) = 3.42 mole
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The chemical formula of the compound can be obtained as follow:

Divide by their molar mass

S = 8.6 / 32 = 0.26875

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Divide by the smallest

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Multiply by 2 to express in whole number

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Thus, the chemical formula is P₂S₅

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12.5 mm = <span>0.0000125 km

1 mm = </span><span>0.00001 hm (hectometer)
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1 mm = </span>0.001 m (meter)
12.5 mm = 0.0125 m

1 mm = 0.1 cm (centimeter)
12.5 mm = 1.25 cm

So the only one of the answer choices that doesn't equal 12.5 mm is 0.00125 hm, since 12.5 mm is <span>0.000125 hm.

Answer:
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</span><span>
Hope this helps!</span>
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