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FinnZ [79.3K]
3 years ago
8

How many chlorine molecules are in 35.5g of chlorine gas

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0
Since chlorine is one of the 7 diatomic elements we know that chlorine appears as Cl₂ gas naturally.  That means that the molar mass of a chlorine gas is 70.9g/mol.  That being said, first you need to find the number of moles of chlorine gas that are present in a 35.5g sample.  To do this divide 35.5g by the molar mass of chlorine gas (70.9g/mol) to get 0.501mol of chlorine.  Then you have to multiply 0.501mol by 6.02×10²³ to get the number of chlorine gas molecules.  Therefore 3.01×10²³ molecules of chlorine gas are present in a 35.5g sample.

I hope that helps.  Let me know in the comments if anything is unclear.
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For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
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Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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Calculate the number of atoms in 5 moles H₂O.
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Answer:

90.35 × 10²³ atoms

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