In the first case, the molarity of oxygen gas is 0.0012 mol/L. In the second case, the molarity of oxygen gas is 0.0052 mol/L .
We have to obtain the number of moles of oxygen as follows;
P = 1.00 atm
V = 28.31 mL or 0.02831 L
n = ?
R = 0.082 atm LK-1mol-1
T = 25 °C + 273 = 298 K
n = PV/RT
n= 1.00 atm × 0.02831 L/ 0.082 atm LK-1mol-1 × 298 K
n = 0.0012 moles
Molarity = Number of moles/Volume = 0.0012 moles/1.00 L = 0.0012 mol/L
For the second case;
P = 4.47 atm
V = 28.31 mL or 0.02831 L
n = ?
R = 0.082 atm LK-1mol-1
T = 25 °C + 273 = 298 K
n = PV/RT
n = 4.47 atm × 0.02831 L/0.082 atm LK-1mol-1 × 298 K
n = 0.1265/24.436
n = 0.0052 moles
Molarity of oxygen = 0.0052 moles/1.00 L = 0.0052 mol/L
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The difference of the blood
To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:
1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>