Ethane is C2H6, with a molar mass of 30g/mol. 60g of ethane is 2mol. Avogadro's number (# of particles in a mole) = 6.02x1023. Avogadro x 2mol = 1.2x1024 ethane molecules.
30 g of ethane will have 6.023 x 1023 molecules of ethane So, 5.5 g will have 1.104 x 1023 molecules of ethane Since there are two carbon atoms, in one molecule of ethane, 5.5 g of ethane will have 2.208 x 1023 atoms of carbon.
To calculate the number of molecules in 15 grams of ethane (C2H6), first find the molar mass of C2H6, which is 30.07 g/mol. Next, calculate the number of moles in 15 grams using the formula: moles = mass / molar mass. Finally, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
The molar mass of acetic acid (C2H4O2) is 60 g/mol. Therefore, there are 1 mole of acetic acid molecules in 60 g. Based on Avogadro's number (6.022 x 10^23 molecules/mol), there are approximately 6.022 x 10^23 molecules in 60 g of acetic acid.
The specific enthalpy of combustion of ethane is approximately -1560 kJ/mol.
The molar volume of a gas at STP (Standard Temperature and Pressure) is 22.4 L. Ethane gas has a molar mass of 30.07 g/mol. Therefore, the mass of ethane gas in a 5.00 L vessel at STP can be calculated as (5.00 L / 22.4 L) * 30.07 g/mol.
30 g of ethane will have 6.023 x 1023 molecules of ethane So, 5.5 g will have 1.104 x 1023 molecules of ethane Since there are two carbon atoms, in one molecule of ethane, 5.5 g of ethane will have 2.208 x 1023 atoms of carbon.
To calculate the number of molecules in 15 grams of ethane (C2H6), first find the molar mass of C2H6, which is 30.07 g/mol. Next, calculate the number of moles in 15 grams using the formula: moles = mass / molar mass. Finally, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
The molar mass of acetic acid (C2H4O2) is 60 g/mol. Therefore, there are 1 mole of acetic acid molecules in 60 g. Based on Avogadro's number (6.022 x 10^23 molecules/mol), there are approximately 6.022 x 10^23 molecules in 60 g of acetic acid.
The heat of vaporization for ethane is approximately 16.7 kJ/g. Therefore, to vaporize 5 g of ethane, the energy required would be: 16.7 kJ/g * 5 g = 83.5 kJ.
1. Given: mass = 0.334g of C2H6, CAtomicWeight = 12.01, HAtomicWeight = 1.008 Required: a. Molar Mass of Ethane b. Number of molecules of C2H6 in 0.334g of C2H6 Formula: a. (n = (2 x CAtomicWeight) + (6 x HAtomicWeight) b. mass x (1 mole / a ) x (6.022x1023 molecules / 1 mole) Solution: a. Molar Mass of Ethane n = (2 x 12.01) + (6 x 1.008) n = 24.02 + 6.0498 n = 30.0698 b. Number of molecules n = 0.334 x (1 mole / 30.0698 g) x (6.022x1023 molecules / 1 mole) n = 6.689x1021 molecules FINAL Answer: The are 6.689x1021 molecules in 0.334g of Ethane
Ethane is C2H6.The answer is 24,23 L water vapors (for a density of 0,804 g/cm3).
Ethane is C2H6.The answer is 24,23 L water vapors (for a density of 0,804 g/cm3).
The specific enthalpy of combustion of ethane is approximately -1560 kJ/mol.
22.0 g of silver chloride contain 0,918.10e23 molecules.
26.3 g of calcium hydroxide contain 2,054 molecules.
Ethanol molar mass: 46,07 gWater molar mass: 18 gAvogadro number: 6,02214129.10 ex.2346,07 g--------------Avogadro number=number of molecules60 g------------------xx = 60 . A/46,07 = 7,842.10 ex.23 molecules of ethanol18 g--------------Avogadro number=number of molecules60 g------------------xx = 60 . A/18 = 20,074.10 ex.23 molecules
32 g SO2 x 1 mole SO2/96 g x 6.02x10^23 molecules/mole = 2.0x10^23 molecules