What mass of h2o is required to form 1.6 l of o2 at a temperature of 325 k and a pressure of 0.995 atm ?

2.14902 g

Explanation:

First thing's first, we have to write out the balanced chemical equation. This is given as;

2H2O --> O2 + 2H2

From the equation above, we can tell that 2 moles of H2O forms 1 mole of O2.

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

This means; 3.2L ( 2 * 1.6) OF H2O would form 1.6L of O2.

Other parameters given are;

Pressure = 0.995atm

Temperature = 325K

Mass = ?

But mass = number of moles * Molar mass

Molar mass of H2O = 18g

Number of moles?

From ideal gas equation;

PV = nRT

n = PV / RT

R = gas constant = 0.082057 L atm mol-1 K-1

n = (0.995 * 3.2) / (0.082057 * 325)

n = 3.184 / 26.669

n = 0.11939 moles

Mass = 0.11939 * 18 = 2.14902 g

The [H3O+] in his stomach is 3,02x10-5 M. See the explanation below, please.

Explanation:

We use the formula:

pH= -log (H30+)

(H30+)= antilog - (pH) =antilog - (4,52) = 3,02e-5 M