pv=nrt. The volume, in litres is calculable by this formula: v=nrt/p, where v is in litres, n is your (1.5x10^25 molecules)/(Avrogadro's 6.022×1023 mol–1), r is a constant with a value of 8.31, t is temperature in Kelvin and p is pressure in Pascals.
A 2-liter bottle of Sprite contains about 5.3 grams of carbon dioxide.
A 12-ounce can of Sprite contains approximately 19 grams of carbon dioxide, which is used to carbonate the beverage.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
To convert from molecules to moles, divide the given number of molecules by Avogadro's number, which is 6.022 x 10^23. Therefore, for 2.22 x 10^23 molecules of carbon dioxide, divide by Avogadro's number to find 0.368 moles of carbon dioxide.
To produce 1 mole of urea, 1 mole of carbon dioxide is needed. The molar mass of urea is 60 grams/mol, and the molar mass of carbon dioxide is 44 grams/mol. Therefore, to produce 125 grams of urea, 125 grams/60 grams/mol = 2.08 moles of urea is needed. This means 2.08 moles of carbon dioxide is needed, which is 2.08 moles * 44 grams/mol = 91.52 grams of carbon dioxide needed.
The gram molecular mass of carbon dioxide is about 44.01 grams. By definition, this value is the number of grams of carbon dioxide that contains Avogadro's Number ("AN") of molecules. Avogadro's Number is about 6.022 X 10^23. Therefore the number of molecules in 1 gram is (1/44.01)(AN) or 2 X 10^21 molecules, to the justified number of significant digits.
To calculate the number of moles of carbon dioxide in 19 grams, divide the given mass by the molar mass of carbon dioxide, which is approximately 44 grams/mol. Therefore, 19 grams of carbon dioxide is equal to 19/44 ≈ 0.43 moles.
1
Six molecules of carbon dioxide are used to produce one 6-carbon sugar molecule through the process of photosynthesis.
A 2-liter bottle of Sprite contains about 5.3 grams of carbon dioxide.
One carbon dioxide molecule is produced in one particle.
Ethane does not have any molecule of carbon dioxide. However when ethane undergoes combustion then two molecules of carbon dioxide are formed (as ethane contains two carbon atoms).
A 12-ounce can of Sprite contains approximately 19 grams of carbon dioxide, which is used to carbonate the beverage.
= 9403.41
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
six molecules
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.