How much space would occupy the CO2 released from a 12 grams CO2 cartridge?
Original question:“How much space would occupy the CO2 released from a 12 grams CO2 cartridge?”Designed 100’s of gas pressure systems, cylinders, and controls & manufactured millions of components from those designs.I will skip the math and simply answer your question:12 grams of CO2 released from a cartridge at sea level and 0 deg C will expand to 6.07 liters or 370.4 co in.We manufactured CO2 cartridges ranging in size from 8 grams to 20 lbs.We also designed the high speed production line for 8 and 12 gram cartridges for the Crosman Arms CO2 pellet gun company.The rule of thumb I developed over 50 years ago is 880 cu in per ounce of CO2 at room temp (70 F)With larger vessels, you need to remember to include the “empty” cylinder as part of your calculations.
12g Co2 cartridge psi?
Pressure can't go above 800 PSI (the vapor pressure of CO2) You will find pressure upto 852 psi @70 deg F,But it is safe to be at 800 PSI.
What is the partial pressure of oxygen and nitrogen in the air if air contains 20% by volume of oxygen and 80% by volume of nitrogen?
partial pressure of O2 =mole fraction of O2 ×total pressure in this case total pressure is 1 atm ( pressure of normal air in atmosphere )mole fraction of O2 = mole of O2 /mole of O2 +mole of n220/ 20+80 ×1= 0.2 atm so partial pressure of O2 =0.2 atmand we know that total pressure = partial pressure of O2 +n21= 0.2 +n 2so partial pressure of nitrogen =1 - o.2=0.8( by Dalton law of partial pressure)I am not sure about answer if it is of heigher class question . my answer is according to class 11
How do you calculate the mass of carbon atoms in CO2?
You use the relative atomic mass of the atoms.Carbon has a relative atomic atomic mass of 12. Oxygen has a relative atomic mass of 16, so two of them is 32.Hence in whichever mass CO₂ you choose, there will be (12 ÷ (12 + 32)) × 100% = 27.3% to 3sf.
Suppose that you have 50 grams of oxygen O2. What is the volume at STP?
First calculate how many moles of O2 you have. O is 16g/mole, so O2 is 32g/mole. 50/32 = 1.5625 moles. 1 mole of any gas at stp is 22.4 liters.1.5625 × 22.4 = 35 liters.Its been 15 years since I graduated HS, and I still remember how to do this. Do your own homework next time and in 15 years you may be able to do the same. ;)
How many air molecules are present in a cubic meter of air?
I am going to have to declare your question as invalid right away. What "air molecules"? There are oxygen molecules, nitrogen molecules, CO2 molecules, even a little bit of hydrogen molecules.Now if you had asked about oxygen & nitrogen molecules, that would have given you pretty close to what you probably want, since the two of them to make up most of air. But I am not going to go through the detailed calculations: I am just going to say that since there must be several moles of both nitrogen and oxygen in a cubic meter, you are going to get a figure well over 10 to the 24th power. That's a lot. No wonder scientists and other scholars believed air was continuous until the late 19th century when Statistical mechanics finally ousted the continuous theory because it predicted thermodynamical properties of gases.
How much volume at STP does 1g of hydrogen occupy?
In many textbooks you must have read the statement,“One mole of any ideal gas occupies 22.4 L at STP.”.Well, not since 1982. The above statement was accurate till 1982 as it was derived from the Ideal Gas Law using the definition of STP till then. In 1982, the definition of STP was slightly modified: Instead of using the pressure as 1 atm or [math]1.01325\times10^5[/math] Pa, we now use exactly [math]10^5[/math] Pa.Now the volume is approximately 22.8 L.Let’s see for ourselves.1g of [math]H[/math] = 1 mol of [math]H[/math] = 0.5 mol of [math]H_2[/math]By the current definition, STP or Standard Temperature and Pressure is,Pressure [math]P = 10^5 Pa = 10^5 \dfrac{N}{m^2}[/math][math] [/math]Temperature [math]T = 273.15 K[/math]Let’s start with the basics;Ideal Gas Law or the Equation of State states that:[math]PV = nRT[/math],where [math]V[/math] is the volume we need to find,[math]n = [/math]the molar count [math]= .5[/math] moland [math]R = [/math]Universal Gas Constant [math]= 8.314 J.K^{-1}.mol^{-1} [/math]Thus we can rewrite the equation of state to[math]V = \dfrac{nRT}{P}[/math]Now substituting the factors by their values we get,[math]V = \dfrac{.5\times(8.314)\times(273.15)}{10^5} m^3[/math][math]\implies V = 0.0114 m^3[/math]As we know that [math]1 m^3 = 10^3 litres[/math] or [math]10^3[/math] L[math]\implies V = 11.4[/math] L (ANS)Thus we can tell that 1 mole of the gas will occupy 22.8 L.P.S. Try using [math]P = 1.01325\times10^5[/math] Pa and see that the volume for 1 mole of gas will be 22.4 L, which is an obsolete value.Your teachers may not have noticed it because they are too busy teaching you science just as religion is taught or as an animal is trained for circus.