Answer is: <span>a. c6h14 and c10h20.
This pair will </span>most likely form a homogeneous solution because they are both nonpolar substances and "li<span>ke dissolves like".
Other pairs will not form homogeneous solution because nonpolar substances have low solubility in polar or ionic substances (for example LiBr is ionic and C</span>₅H₁₂ is nonpolar).
The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
Answer:
0.6 moles NH₃
Explanation:
The reaction that takes place is:
First we <u>determine the limiting reactant</u>:
- 0.35 mol N₂ would react completely with (3*0.35) 1.05 moles of H₂. There are not as many H₂ moles, so H₂ is the limiting reactant.
Then we <u>convert H₂ moles (the limiting reactant) to NH₃ moles</u>, keeping in mind the <em>stoichiometry of the reaction</em>:
- 0.90 mol H₂ *
= 0.6 moles NH₃
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
There are 0.000076 moles in 4.6 x 10^19 atoms.
