General Chemistry Answers

Complete the following for aqueous solutions of Aluminum nitrate. Show Complete Solution.

1. Mass of Solute= 2.544g, Molarity= 1.688M, Volume of Solution=____
2. Mass of Solute= _____, Molarity= 0.729M, Volume of Solution= 894mL

Freon in air conditioning unit has a volume of 0.30L. It is allowed to function in a room, where the pressure is about 628 mm of Hg, assuming that the temperature is in a constant state. Find the final pressure of Freon when its volume is increased to 0.95L.

What will happen to a spray can of paint containing only the propellant at a pressure of 750torr at 30°C if it is thrown into a heap of garbage burning at 165°C?

A 132mL of gas is measured at 38°C. If the pressure remains constant, what will be the volume of the gas at 10°C?

What is the molality of a solution of 2.50g H3PO4 dissolved in 240g water? Show complete solution.

Calculate the Molarity of each of the following solutions:

a. 185 g of sucrose, C12H22011 in 1.00 L of solution

b. 6.30 g of HNO3 dissolved in 255 ml of solution.

A saline solution is prepared by dissolving 5.04 g of NaCl in 95.0 g of water. Calculate the following:

a. Mole fraction of the solute and the solvent b. % mass of the solute

c. Concentration of solution in ppm

Show complete solution.

$$\ce{2C_4H_{10}(g) +13O_2(g) -> 8 CO_2(g) + 10H_2O(l)}$$
$65cm^3$ of butane is burnt in $65cm^3$ of oxygen. Find the volume of carbon dioxide produced.\

I would like someone to explain this to me.

$$\ce{2C_4H_{10}(g) +13O_2(g) -> 8 CO_2(g) + 10H_2O(l)}$$

$65cm^3$ of butane is burnt in $65cm^3$ of oxygen. Find the volume of carbon dioxide produced.\

Solution\

Assume\

i) $O_2$ and $CO_2$ are ideal gas at STP condition $22.7dm^3/mol$\

ii) $C_4H_{10}$ is limiting reactant with lowest coefficient in the equation while $O_2$ is the excess reactant.

$n_{O_2 (real)}=0.065/22.7=2.8634\times10^-3$ mol $O_2$(real)

$$From\qquad equation:$$$$\ce{2C_4H_{10}(g) :13O_2(g) }$$

$$Real:$$$$\ce{X C_4H_{10}(g) :(2.8634\times{10^-3}) O_2(g) }$$

Ratio: $\frac{2}{13}=\frac{X}{2.8634\times{10^-3}}$\

$X=\frac{2}{13}\times(2.8634\times{10^-3})=4.4052\times10^-4$ mol $C_4H_{10}(real)$ which is the limiting reactant.

$$From\qquad equation:$$$$\ce{2C_4H_{10}(g) :8CO_2(g) }$$

$$Real:$$$$\ce{4.4052\times10^{-4} C_4H_{10}(g) :Y CO_2(g) }$$

$Y=\frac{8}{2}\times(4.4052\times10^{-4})=1.7621\times10^-3$ mol $CO_2(real)$ 

$V_{CO_2}(real)=1.7621\times10^-3$ mol $\times 22.7dm^3/mol=0.039999dm^3=0.04dm^3=40cm^3$

Correct?

30% HCI solution by mass