Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL
v/v% = 16.0 / 155 x 100% = 10.32%
this means that in a volume of 100 mL solution, 10.32 mL is acetone.
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
The correct Lewis structure of SO2 is the Lewis structure that shows all the 12 valence electrons in the molecule.
A Lewis structure shows the number of valence electrons on the valence shell of all the atoms in a compound. The electrons are shown as dots around the symbol of each element or a dash to indicate shared electrons in a covalent bond.
Looking at the Lewis structure of SO2 attached to this answer, we can see the twelve valence electrons in the molecule and how they are distributed around each atom as shown.
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Maalox is the trade name for an antacid and antigas medication used for relief of heartburn, bloating, and acid indigestion in which
4 ml contains
= 320mg of aluminum hydroxide
= 320mg of magnesium hydroxide
= 32mg of simethicone
recommended doses = 4 times * 2 tea spoon = 8 tea spoon/ day
given = 1 tea spoon = 5 ml
8 tea spoon = 40 ml
hence,
amount of aluminum hydroxide = 320/4 * 40 = 3200mg = 3.2 g
amount of magnesium hydroxide = 320/4 * 40 = 3.2 g
amount of simethicone = 32/4 * 40 = 320 mg = 0.32g
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