Answer:
A. The partial pressure for CH4 = 0.0925atm
B. The partial pressure for C2H6 = 0.925atm
C. The partial pressure for C3H8 = 0.346atm
D. The partial pressure for C4H10 = 0.115atm
Explanation:
Total pressure = 1.48atm
Total mole = 0.4+4+1.5+0.5=6.4
A. Mole fraction of CH4 = 0.4/6.4 = 0.0625
The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm
B. Mole fraction of C2H6 = 4/6.4 = 0.625
The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm
C. Mole fraction of C3H8 = 1.5/6.4 = 0.234
The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm
D. Mole fraction of C4H10 = 0.5/6.4 = 0.078
The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm
Im not really sure what your asking.... <span>Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, </span>14.70 pounds per square inch<span>, 1,013.25 × 10 </span>3<span> dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.
</span><span>""atmospheric pressure | Britannica.com""</span>
Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.
<span>0.453 moles O2 is the solution</span>
Answer:
1.1 × 10² g
Explanation:
First, we will convert 1.0 L to cubic centimeters.
1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³
The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:
1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g
1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:
1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g
