GAS LAW and other PRE LAB HELP

[ANIMAL who]

#1-Use the Ideal Gas Law to calculate the number of moles of gas at 1.00atm in a cell 10.0cm long having a cross sectional area of 1.00cm^2 at 298K. (1ml=1cm^3)


#2-Lets say you start with 6.0 x 10^2 mL of 1.0 x 10^-4 M NaOH solution containing phenolphthalein indicator. The equation is CO_2 + OH^- yields HCO^-_3 (HCO bottom number of three with an overall negative charge) If 374 mL of a persons breath is required to turn the phenolphthalein from pink to colorless, calculate the number of moles of CO_2 per L of breath (ie., the concentration of CO_2) in this breath.

 

[chemisttree]

#1-Use the Ideal Gas Law to calculate the number of moles of gas at 1.00atm in a cell 10.0cm long having a cross sectional area of 1.00cm^2 at 298K. (1ml=1cm^3)


#2-Lets say you start with 6.0 x 10^2 mL of 1.0 x 10^-4 M NaOH solution containing phenolphthalein indicator. The equation is CO_2 + OH^- yields HCO^-_3 (HCO bottom number of three with an overall negative charge) If 374 mL of a persons breath is required to turn the phenolphthalein from pink to colorless, calculate the number of moles of CO_2 per L of breath (ie., the concentration of CO_2) in this breath.

Show us what you have so far, Animal.

 

[Blueshift5]

I would like to provide a response to the content regarding the Ideal Gas Law and the pre-lab help needed.

Firstly, to calculate the number of moles of gas at 1.00atm in a cell 10.0cm long with a cross-sectional area of 1.00cm^2 at 298K, we can use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. In this case, we are given the pressure (1.00atm), volume (10.0cm x 1.00cm^2 = 10.0cm^3 = 10.0mL), and temperature (298K). We can rearrange the equation to solve for n, which gives us n = PV/RT. Plugging in the values, we get n = (1.00atm x 10.0mL)/(0.0821 L atm/mol K x 298K) = 0.00425 moles of gas.

Moving on to the second part, we have been given a NaOH solution with a concentration of 1.0 x 10^-4 M and phenolphthalein indicator. The equation given is CO_2 + OH^- yields HCO^-_3, which suggests that CO_2 is reacting with OH^- to form HCO^-_3. The question asks us to calculate the number of moles of CO_2 per L of breath, which is essentially the concentration of CO_2 in the breath.

To find the concentration, we can use the formula C = n/V, where C is the concentration, n is the number of moles, and V is the volume. We have been given the volume of breath (374mL) and the number of moles of NaOH (6.0 x 10^2 mL x 1.0 x 10^-4 M = 0.06 moles). However, we need to convert the volume of breath from mL to L, which gives us 0.374L. Plugging in the values, we get C = 0.06 moles/0.374L = 0.16 M. This means that the concentration of CO_2 in the