Answer:
mass of excessive CO = 2.55 gram
Explanation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
moles of Fe2O3 = mass / formula mass = 22.00/(56x2+16x3)=0.1375 (mol)
moles of CO = mass / formula mass = 14.1/(16+12) = 0.503
Fe2O3 reacts completely meanwhile CO is excessive.
mass of CO reacts = 3 x nFe2O3 x M = 3 x 0.1375 x (16+12) = 11.55 gram
mass of excessive CO = initial mass - reacted mass = 14.1 - 11.55 = 2.55 gram
The correct answer is (C) in through the pores and out through the osculum
Answer: The percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.
Explanation:
Given: Volume of solute = 120 mL
Volume of solvent = 350 mL
Now, total volume of the solution is as follows.

Let us assume that 100 mL of solution is taken and the amount of isopropyl alcohol present in it is as follows.

Hence, there is 25.53 mL isopropyl alcohol is present in 100 mL of solution. Therefore, %v/v is calculated as follows.

Thus, we can conclude that the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.