4. What is the temperature in levels C that 5.78 together of a gas in ~ 735mmHg if it rectal 6.19 L at 18 degrees C and 925 mmHg?

Itlooks prefer this difficulty involves pressure, volume and also temperature (P,V,T), sowe require the general gas legislation (GGL): . We deserve to rearrange it to solve for the last temperature, T2: 5. How numerous moles of gas are present in a volume that 378 mL in ~ 710 mmHg and also -22 degrees C?

Hereis a question that entails one of the big THREE (number of moles, a density ora number of molecules). In this case, it entails moles. So that meanswe need to use the IGL (ideal gas law) i m sorry relates pressure, volume,temperature and also amount (number of moles). The IGL is, that course, PV = nRT where n is the number of moles and also R is a proportionality continuous called theUniversal Gas Constant. R deserve to have values of 0.08206 or 62.4 depending upon the devices you want for her answer or the unitsgiven in your problem. So... Back to our problem. The question asks "How manymoles..." therefore let"s settle the IGL for n: Temperature has tobe convert to Kelvins: -22 levels C+273.15 = 251K. Weconvert 378 mL to liters which is 0.378 L. We have achoice of just how to treat the pressure. It"s provided in mmHg, so we can use the62.4 worth for R and also use the mmHg pressure directly: Notice how 378 mL to be cleverly changed to 0.378L and degrees C was adjusted to Kelvins on-the-fly? us could additionally use the 0.08206 worth for R by an initial converting the pressure from mmHg to atm. Due to the fact that 1 atm = 760 mmHg(exactly), us can easily convert:  6. What is the push of 6.54 mL of a gas sample at 57 levels C if it rectal 9.75 mL at21 levels C and 1.64 atm?

Thislooks like another PVT problem. Therefore we usage . Therefore we select which collection of PVT data has actually the subscripts 1 andwhich have the subscripts 2. P1 = 1.64 atm,V1 = 0.00975 L, T1 = (21 levels C+273.15). So, V2 = 0.00654L, T2 = (57 degrees C+273.15).

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STP method T = 273Kand ns = 1 atm. The thickness of a gas have the right to be calculatedfrom the IGL (Ideal Gas Law) since density is just mass every unit volume. If wetake n, the variety of moles, we can transform it to grams by making use of M, the molarmass: m = n´M which method . V is currently in the IGL (PV = nRT),so we substitute for n: So currently we have anexpression for density in terms we know. If m is in grams and V is in liters,the thickness will be in g/L. 8. What is the volume the 216.8 g of oxygen at STP?

STP means T = 273Kand p = 1 atm. Since grams of oxygen space involved, itis clear the we need to use the IGL. We deserve to use the 216.8 grams and also the molarmass of oxygen to calculate n, the variety of moles. STP gives us T and P. R isa continuous that we know. Therefore we"re every set.

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From trouble 7, weshowed the . So, offered the density and also STP, we should be able to find M,the molar mass. 10. Whatis the press of 2.50 moles of oxygen if the sample lived in 40.0 L at 25 levels C?

Aha! The problemmentions moles! for this reason we understand it requires the use of the IGL. PV = nRT. Rearranging to offer pressure,