For this problem you have to use PV=nRT.
In order to get anywhere with the question, you have to isolate the V (volume) and get it alone. To do that, you have to divide each side by P (pressure). When you do that, your new formula to work with is:
V = nRT / P
(the P cancelled out on the other side, which left you with V)
Unfortunately, you weren't given the number of moles in the problem. Instead, they made it a bit more challenging by giving you the number of grams, which can be converted to give you your number of moles:
42 g / 1 ( 1 mole / 28.01 g) = 1.5 moles of CO
Now, you have all the information that you need in order to answer the question. All you have to do now is plug in, but make sure that you know where everything goes:
n (moles) = 1.5 moles
R = .0821 L atm / mole K <-----that one looks weird, but that's all required for R
T (temp) = 273 K <----that's your standard temp *must be in Kelvin!!!! ALWAYS!!
P (pressure) = 1 atm <----standard pressure for P *must be in atm!! ALWAYS!!
Your equation should now look like this:
V= (1.5 moles x .0821 Latm/ moleK x 273 K) / (1 atm) =
34 L CO
And then you get your answer!
Vstp = ( n ) ( 22.42 L/mol ) = ( m/M ) ( 22.42 L/mol )
Vstp = ( 92.31 g / 44.01 g per mol ) ( 22.42 L per mol at STP )
Vstp = ( 2.097 mol ) ( 22.42 L/mol )
Vstp = 47.03 L <-----[ Answer ]
The volume is 46,69 L.
The volume is 66,77 mL.
2.24 L
8.96 cm3
The volume of CO2 is 53,18 litres.
Atomic mass of C = 14g/mol Atomic mass of O = 16g/mol Molecular mass of CO2 = 12 + 2(16) = 44g/mol mass = number of moles x molecular mass mass = 3 mol x 44g/mol = 132g
61ml
22.4 liters at STP
CaCO3 + 2HCl -> CaCl2 + CO2 +H2O
The volume of CO2 is 53,18 litres.
Auto emits carbon monoxide(CO), carbon dioxide(CO2) and some other gases and particles which pollute the environment. the amount of CO2 compared to total amount of emission in percentage is called as percent of carbon dioxide in auto emissions. %CO2=(Volume of CO2 emitted/Total volume of emission)*100
CO2(g)
Atomic mass of C = 14g/mol Atomic mass of O = 16g/mol Molecular mass of CO2 = 12 + 2(16) = 44g/mol mass = number of moles x molecular mass mass = 3 mol x 44g/mol = 132g
CO2 floats because its density is less then water. Anything will float if its density is less then water. That is; when a certain volume of CO2 (or anything else) weighs less then the same volume of water.
61ml
5126 cm3
22.4 liters at STP
CaCO3 + 2HCl -> CaCl2 + CO2 +H2O
The molar volume is 22,414 L at 0 oC (this is a law in chemistry).22,414 x 65 = 1 457 L CO2
The volume is 1,1 mL.
Volume is proportional to temperature, so if there is any increase in temperature, the volume of the gas will increase proportionally