The question is incomplete, here is the complete question:
A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?
<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL
<u>Explanation:</u>
We are given:
Volume of bottle of rubbing alcohol = 946 mL
70% (v/v) alcohol solution
This means that 70 mL of rubbing alcohol is present in 100 mL of solution
Amount of water present in solution = [100 - 70] = 30 mL
Applying unitary method:
In 100 mL of solution, the amount of water present is 30 mL
So, in 946 mL of solution, the amount of water present will be = 
Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL
Answer:
3 valence electrons
Explanation:
Because aluminum has three, that means three fluorines can bond. The make the formula AlF3, also known as aluminum trifluoride. Each of the fluorine atoms gets an electron to fill their shell, and the aluminum loses three, giving it a filled shell too (remember, aluminum has three extra electrons).
Answer:
4.79 g of water
Explanation:
From the reaction equation;
C2H6(g) + 7/2O2(g) ----> 2CO2(g) + 3H2O(g)
Next we convert the given masses of reactants to moles of reactants.
For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles
For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles
Next we determine the limiting reactant, the limiting reactant yields the least amount of product.
For ethane;
From the reaction equation,
1 mole of ethane yields 3 moles of water
0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water
For oxygen;
3.5 moles of oxygen yields 3 moles of water
0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water
Hence oxygen is the limiting reactant.
Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water
