Number of moles = mass of product / molecular mass
=mass of product (MgO) / 40.3
Since the mass of MgO is not given in the question, the correct answer choice cannot be given. However, proceeding witht eh above formula will enable you to find the correct number of moles given the mass of MgO.
Density * Volume = Mass
Now we substitute the values in.
19.3 g/cm^3 + 20 cm^3 = 386 g
Mass = 386 g
Answer:
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Explanation:
<span>Answer:
</span><span>
</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />
<span>Explanation:
</span>
<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
</span>
<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>
<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
</span>
<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>
<span>3) Finally, you can wirte the total ionic equation:
</span>
Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
