Q: How many grams are there in 1.5 x 1025 molecules of CO2?

Write your answer...

Submit

Still have questions?

Related questions

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.

For this you need the atomic (molecular) mass of CO2. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. CO2=44.0 grams33.0 grams C / (44.0 grams) = .750 moles CO2

CO2's molecular weight is 44 so 44 g of CO2 contain Avogadro's number of CO2 molecules. 20 g would contain 20/44 x Avogadro's number.

One mole is defined by Avogadro's number of 6. 02x10^23 particles. To solve this equation we multiple Avogadro's number by our given quantity of 2. 10 moles. The answer then is 2. 64x10^24 molecules of CO2.

6.5 grams CO2 divided by 44 grams CO2 per mole CO2 = 6.8 mole CO2 (Molar mass CO2 = 12 + 2*16 = 44 g/mol)

To find the mass of 3.5 x 10^22 molecules of CO2, we need to multiply the number of molecules by the molar mass of CO2. The molar mass of CO2 is approximately 44 grams per mole. So, 3.5 x 10^22 molecules of CO2 would have a mass of approximately 1.54 x 10^25 grams.

8.066

The number of molecules is 0,90332112855.10e23.

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.

The balanced equation for the combustion of CH4 is CH4 + 2O2 ==> CO2 + 2H2O4 molecules of CH4 will produce 4 molecules of CO2 and 8 molecules of H2O

Each mole of a substance contains 6.022 E23 molecules or atoms of that substance. Four moles of H2O will contain 2.4088 E24 molecules.

92.4 grams