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zysi [14]
3 years ago
7

How many moles of chlorine are in 6.67x10^40 chlorine molecules?

Chemistry
2 answers:
Vladimir [108]3 years ago
6 0

To determine the moles of chlorine atoms you would first need to multiply your number by 2 and then divide that by 6.022 x 10^23 (Avogadro’s number). If you want to determine the number of moles of chlorine (Cl2), then you would divide that result by 2. So:

1.1076 x 10^17 moles of chlorine (Cl2)

2.2152 x 10^17 moles of chlorine (Cl) atoms.

sweet-ann [11.9K]3 years ago
4 0
To solve for the number of moles, we simply have to use the Avogadros number which states that there are 6.022 x 10^23 molecules per mole. Therefore:
 
number of moles = 6.67 X 10^40 chlorine molecules / (6.022 x 10^23 molecules / mole)
number of moles = 1.108 x 10^17 moles



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How many atoms of potassium are in 0.250 mol of potassium carbonate, k2co3?
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The  number  of  potassium  atom  that  are  in  0.25  moles   potassium  carbonate  is  calculated  as  follows

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4 0
3 years ago
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

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At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

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7 0
3 years ago
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