A sample of oxygen gas has a volume of 5.60 l at 27Â°c and 800.0 torr. how many oxygen molecules does it contain?

1 answer:

8 0

<span>1.44x10^23 molecules of oxygen gas The ideal gas law is PV = nRT where P = pressure (800.0 Torr) V = volume (5.60 L) n = number of moles R = Ideal gas constant (62.363577 L*Torr/(K*mol) ) T = absolute temperature (27C + 273.15 = 300.15 K) Let's solve for n, the substitute the known values and solve. PV = nRT PV/RT = n (800.0 Torr*5.60 L)/(62.363577 L*Torr/(K*mol)*300.15 K) = n (4480 L*Torr)/(18718.42764 L*Torr/mol) = n 0.239336342 mol = n So we have 0.239336342 moles of oxygen molecules. To get the number of atoms, we need to multiply by avogadro's number, so: 0.239336342 * 6.0221409x10^23 = 1.44x10^23</span>

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timama [110]

To be able to calculate the number of moles for this problem, for simplicity, we assume that it is an ideal gas. We use the equation PV = nRT. We do as follows:

PV = nRT
n = PV / RT
n = 1(100000) / 0.08206 (27 + 273.15)
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8 0

4 years ago

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The side chain of histidine has a pKa of about 6. What percent of histidine side chains would be deprotonated at pH 7.5

Setler [38]

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