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Veseljchak [2.6K]
3 years ago
8

12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)

Chemistry
1 answer:
Sergeu [11.5K]3 years ago
4 0
2NaClO₃   →  2NaCl  +  3O₂

mole ratio of NaClO₃  to  O₂  is  2  :  3

∴  if moles of NaClO₃  =  12 mol

then moles of O₂  =  \frac{12 mol   *   3}{2}
      
                             =  18 mol


Mass of O₂  =  mol of O₂  ×  molar mass of O₂
 
                    =  18 mol  × 16 g/mol
 
                    =  288 g

So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction.  If you should have used another reaction then just follow the same steps and you'll get your answer.




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3 years ago
What is the predominant intermolecular force in the liquid state of each of these compounds: ammonia (nh3), carbon tetrachloride
Kay [80]
  • NH₃: Hydrogen bonds;
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<h3>Explanation</h3>

Relative strength of intermolecular forces in small molecules:

Hydrogen bonds > Dipole-dipole interactions > London DIspersion Forces.

It takes two conditions for molecules in a substance to form <em>hydrogen bonds</em>.

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  • They shall contain at least one lone pair of electrons.

NH₃ contains N-H bonds. The central nitrogen atom in an NH₃ molecule has one lone pair of electrons. NH₃ meets both conditions; it is capable of forming hydrogen bonds.

CCl₄ molecules are nonpolar. The molecule has a tetrahedral geometry. Dipole from the polar C-Cl bonds cancel out due to symmetry. The molecule is nonpolar overall. As a result, only London Dispersion Force is possible between CCl₄ molecules.

HCl molecules are polar. The H-Cl bond is fairly polar. The HCl molecule is asymmetric, such that the dipole won't cancel out. The molecule is overall polar. Both dipole-dipole interactions and London Dispersion Force are possible between HCl molecules. However, dipole-dipole interactions are most predominant among the two.

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?SiCl4(I) + ?H2O(I) = ?SiO2(S) + ?HCI(aq)​
docker41 [41]
SiCl4 + 2H2O = SiO2 + 4HCl
5 0
3 years ago
Read 2 more answers
Calculate the molarity of a 150 g NaCl in 250 ml solution.
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Answer: 10.3 mol/l

Explanation: Molality = concentration c = n/V

Amount of substance n = m/M = 150 g / 58.44 g/mol  =

2,5667 mol . c = 2.5667 mol / 0.25 l = 10.2669 mol/l

c =

7 0
3 years ago
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