This is given by Avogagro number: 1 mol = 6.02*10^23 particles
Then you can do whichever to these two relations, because they are equivalent:
- 1mol / 6.02*10^23 representative particles, and
- 6.02*10^23 representative particle /1 mol
Only the second option of the question includes one of the valid conversion factors. Then, the conversion factor of the second option is the right answer
True. Each contour line on a topographic map represent a certain elevation; the closer they are, the steeper the surface is. All points on a contour line are the same elevation.
Answer: Friction is a force that resists relative motion between two surfaces in contact. Depending on the application, friction may be desirable or undesirable.
Answer: The molality of solution is 17.6 mole/kg
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
where,
n = moles of solute
= weight of solvent in kg
moles of acetone (solute) = 0.241
moles of water (solvent )= (1-0.241) = 0.759
mass of water (solvent )= 
Now put all the given values in the formula of molality, we get
Therefore, the molality of solution is 17.6 mole/kg
Answer:
Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>
- It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
- Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
- The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.
- <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>
∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.
- From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
- So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
- Finally, we can calculate the mass of copper produced using:
mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.
- <u><em>So, the answer is:</em></u>
<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>
