Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
Answer: 1.09 g
Explanation:
If we use the approximation that 1 mole is 22.4 L, then setting up a proportion,
- 1/22.4 = x/0.345 (x is the number of moles in the sample)
- x = 0.0154 mol
Since the mass of a mole of chlroine is about 70.9 g/mol, (0.0154)(70.9) = 1.09 g (to 3 s.f.)
Answer:
Explanation:
Ethanol is the solvent and sucrose is the solute
If 6 units of AB were reacted with 10 units of CD in the described equation, the limiting reactant would AB
<h3>Limiting reactants</h3>
They are reactants that determine how far a reaction can go in terms of yield.
From the equation: 3AB + 4CD --> 2AD + 6CB
The mole ratio of AB to CD is 3:4
Thus, 6 units of AB will require 8 units of CD.
But 10 units of CD were reacted with only 6 units of AB. This means that CD is in excess by 2 units while AB will limit the yield of the reaction.
More on limiting reactants can be found here: brainly.com/question/14225536
<h2>Antacid is a common acid</h2>
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