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According to "Newton's First Law" an object at rest will begin to move when it is acted upon an unbalanced force.
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Soil temperature and <span>water content</span>
Answer:
Divide the mass of your anhydrous (heated) salt sample by the molar mass of the anhydrous compound to get the number of moles of compound present. In our example, 16 grams / 160 grams per mole = 0.1 moles. Divide the mass of water lost when you heated the salt by the molar mass of water, roughly 18 grams per mole.In order to determine the formula of the hydrate, [Anhydrous Solid⋅xH2O], the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid (Equation 2.12. 6).
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>
Answer:
negative
the chage on hydroxide ions os negative
