<span>34.2 grams
Lookup the atomic weights of the involved elements
Atomic weight potassium = 39.0983
Atomic weight Chlorine = 35.453
Atomic weight Oxygen = 15.999
Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol
Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol
The balanced equation for heating KClO3 is
2 KClO3 = 2 KCl + 3 O2
So 2 moles of KClO3 will break down into 3 moles of oxygen molecules.
0.713188188 mol / 2 * 3 = 1.069782282 mols
So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be
1.069782282 * 2 * 15.999 = 34.23089345 grams
Rounding the results to 3 significant figures gives 34.2 grams</span>
Answer:
has boiling point of 238 K
Explanation:
Boiling point depends on different intermolecular force such as molecular wight, dipole-dipole attraction force, hydrogen bonding, ionic attraction force.
Homonuclear diatomic molecules are covalent non-polar molecules and thereby free from dipole-dipole attraction force, hydrogen bonding and ionic interaction forces.
Hence, boiling point of homonuclear diatomic molecules depends solely on molecular weight.
We know, higher the molecular weight of a molecule, higher will be its boiling point. This phenomenon can be realized in terms of increasing london dispersion force with increase in molecular weight.
Decreasing order of molecular weight of halogen molecules :
>
>
>
So, decresing order of boiling point of halogen molecules:
>
>
>
Hence
has boiling point of 238 K
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L