== == == == Standard temperature and pressure (STP), as defined by IUPAC, is 0 degrees Celsius (273 Kelvin) and 1 atmosphere of pressure.
Use the ideal gas law:
P V = n R T
where P is pressure in atmospheres, V is volume in liters, n is moles of gas, R is the universal gas constant, and T is temperature in Kelvin.
(1) (44.8) = n (0.0821) (273)
Substitute values in, and solve for moles.
n = approximately 1.9988132 moles.
With this new information, we can get our answer.
Multiply moles by oxygen's molar mass. In our case, this value is 32, as oxygen gas exists in diatomic molecules (meaning there's two oxygen atoms per molecule).
1.9988132 * 32 = 63.9620224 grams of oxygen.
Here is your answer. Apply significant figures as necessary. Look at the related question, "How do you solve Ideal Gas Law problems?" for more information. Credit for that answer goes to JEK, as of this writing.
I forget the easier way, so we do it this way. Use PV = nRT and find moles O2.
(1 atm)(44.8 liters) = n(0.08206 L*atm/mol *K)(298.15 K)
= 1.831 moles O2
1.831 moles O2 (32.0 grams/1 mole O2)
= 58.6 grams of O2
That's certainly going to depend on what substances and how many of each
are present inside the container before anything forms, and on what kind of
chemical processes take place in there.
4.4 gm of CO2 is 0.1 mole.
1 mole CO2 contains 1 mole O2 molecules.
Therefore, 0.1 mole contains 0.1 mole of O2 molecules i.e. 6.022 x 1022 molecules of Oxygen
134,4 g oxygen are needed.
2.4g
5.6 moles
31.1 gm
28.35 gm in 1 ounce. So 550 gm is 19.4 ounces
190mmh x510mm
2304 gm = 2304/1000 [ kg ] = 2304/1000 x 2.2 [ pounds ]
0.25moles
you have 3 on the astro & safari vans 4.3L
A gram-mole of any gas occupies 22.4 litres. The molecular weight of CO2 is one of carbon plus two of oxygen, 1 x 12 + 2 x 16, in other words 44. That means that a gram-mole of CO2 is 44 grams, which occupies 22.4 litres or 22,400 cc, so its density is 44/22400 or 0.00196 gm/cc. The MW of hydrogen is 2, nitrogen is 18, oxygen is 32, so CO2 is a relatively dense gas.
If you mean o2 sensors, there is one. It is located on the exhaust manifold on the back side of the engine. Gm started the two o2 sensors around 94-95 when obdII was adopted.
4.50 moles NO2 X (46 grams NO2) / (1 mole NO2) x (4 moles NO2) / (7 moles O2) x (1 mole O2) / (32 gm O2) = 3.70 grams O2
So CO2 is 44g/mol. 48/44 is 1.0909091 moles. This multiplied by avagadros number is 6.57x1023 molecules. As there is 1 carbon per molecule, this also equals the number of carbon atoms
if code 44 is a lean code and 45 is a rich code if both are present suspect faulty o2 sensor if just code 44 check fuel pressure it should never drop below 10psi. then block off the return line from the throttle body. pressure should raise to +18psi. if not replace fuel pump.
O2 sensor CKT slow response (bank 1 sensor 1). Sounds like a tired B1S1 O2 sensor if all the connections are good.
1000grams in a kilogram.
This discussion will contain approximate values. Plug in the real values for a more accurate answer. One litre of petrol (gasoline for American readers) weighs about 800 grams. It is composed principally of carbon and hydrogen with a ratio of 2 carbon per 2 (and a bit) hydrogen or 24 units of mass carbon per 2 units of mass hydrogen. This means that for every 800 gm (one litre) of fuel you have (24/26)x800 = 738 grams of carbon. Since 1 carbon atom combines with 2 oxygen atoms each 12 gm carbon combines with 32 gm oxygen to make 44 gm Carbon dioxide. So for every litre of fuel (which weighs 800gm) you have 738 gm carbon and require 1969 grams of oxygen to combust it. This produces 738+1969= 2707 grams of CO2.
All O2 sensors are threaded into the exhaust system at various points.
on the drivers side underneath the engine by the front tire. You will notice a pigtail leading to the exhaust in this area. This o2 sensor is the main sensor, there should be a catalytic o2 sensor located after the converter.
631.86 gm equals how many ounces in cocaine ?