Answer:
I can't answer this as I am not able to read the Question
In a 2.2 m solution with the same volume there would be 11.825 g of a salt
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
Hi my name is Jaleesa The answer = -1
Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.
This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.
Next, let's determine how many moles of butane are burned. This is obtained by
5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.
Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.
We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).
V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.
Explanation:
