0
Add your answer:
Earn +20 pts
How many grams of co2 are in 2l mol of the compound?
92.4 grams
How many grams of CO2 are produced from 5 moles O2?
5.0 grams CO2 (1mol CO2/44.01g) = 0.11 moles CO2
How many moles are there in 6.5 grams of CO2?
To find the number of moles in 6.5 grams of CO2, you first need to calculate the molar mass of CO2, which is 44.01 g/mol. Then, divide the given mass by the molar mass to get the number of moles. In this case, 6.5 grams / 44.01 g/mol = 0.148 moles of CO2.
How many grams are there in 1.5 x 1025 molecules of CO2?
1100 Grams. or 2.42 pounds.
How many grams of CO2 are in 2.1 mol of the compound C and O?
I think 92.4 grams
When 125 grams of C2H4 burns in oxygen to give carbon dioxide and water how many grams of CO2 are formed?
Balanced equation C2H4 + 3O2 -> 2CO2 + 2H2O 125 grams ethylene (1 mole C2H4/28.052 grams)(2 mole CO2/1 mole C2H4)(44.01 grams/1 moleCO2) = 392 grams CO2 formed ==================
How many grams of CO2 will be produced by burning 22 gram of C3H8?
To calculate the grams of CO2 produced by burning 22 grams of C3H8, first determine the moles of C3H8 using its molar mass, then use the balanced chemical equation for the combustion of C3H8 to find the moles of CO2 produced, and finally convert moles of CO2 to grams using the molar mass of CO2.
How many grams of CO2 are produced in 2.09 mol of HCl?
CaC12
What is the mass of a half a mole of CO2?
The molar mass of CO2 is approximately 44.01 g/mol. Therefore, half a mole of CO2 would have a mass of around 22.0 g.
How many grams of CO2 will be produced by burning 34.3 gram of C3H8?
To calculate the amount of CO2 produced when burning 34.3 grams of C3H8 (propane), you need to balance the chemical equation for the combustion of C3H8. Since each mole of C3H8 produces 3 moles of CO2, you first convert 34.3 grams of C3H8 to moles, calculate the moles of CO2 produced, and then convert that to grams of CO2.
How many grams of CO2 are in 2.1 mol of the compound?
99
How many grams of CO2 are in 2.1 mol of compound?
92.4- Nova Net
