To calculate the mass of four moles of methane (CH4), we first need to determine the molar mass of methane. The molar mass of CH4 is the sum of the atomic masses of carbon (12.01 g/mol) and hydrogen (1.008 g/mol) multiplied by the number of atoms in the compound. Therefore, the molar mass of CH4 is 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol. To find the mass of four moles of methane, we multiply the molar mass by the number of moles: 16.04 g/mol x 4 mol = 64.16 grams. Therefore, four moles of methane would have a mass of 64.16 grams.
The molar mass of methane (CH4) is 16 g/mol. Therefore, four moles of methane would have a mass of 64 grams.
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
The molar mass of NH3 is 17.03 g/mol. To find the mass of 3 moles of NH3, you would multiply the molar mass by 3. Therefore, the mass of 3 moles of ammonia would be 51.09 grams.
The molar mass of gold is approximately 197 grams/mol. To convert moles to grams, you multiply the number of moles by the molar mass. Therefore, 2.8 moles of gold would be approximately 2.8 * 197 = 551.6 grams.
The molar mass of methane (CH4) is 16 g/mol. Therefore, four moles of methane would have a mass of 64 grams.
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The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
16.0 grams of methane (CH4) is equivalent to about 0.92 moles of methane, since the molar mass of methane is approximately 16.04 g/mol. In terms of molecules, this would be approximately 5.53 x 10^22 molecules of methane.
The mass would be approximately 92 grams.
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
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The amount of grams in two moles depends on the element or compound that you have two moles of. For example, the molar mass of oxygen is about 16 grams. So two moles of oxygen would be about 32 grams.
The molar mass of NH3 is 17.03 g/mol. To find the mass of 3 moles of NH3, you would multiply the molar mass by 3. Therefore, the mass of 3 moles of ammonia would be 51.09 grams.
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
You can only calculate the empirical formula because you do not have a mass of this compound given. To do the empirical formula assume 100 grams and change percent to grams. Get moles. 80 grams Carbon (1 mole C/12.01 grams) = 6.66 moles C 20 grams hydrogen (1 mole H/1.008 grams) = 19.84 moles H the smallest becomes 1 in the empirical formula and the other number is divided by it, Thus; H/C 19.84 moles H/6.66 moles C = 2.9, which we call 3 so, CH3 --------------- is the empirical formula To get the molecular formula tour question needed to read; How to calculate molecular formula from such ans such mass of compound with these percentages of elements, Which, of course, your question did not provide. Then you would have divided that given mass by the mass total of the elements of the empirical formula, got a whole number by which you would have multiplied the numbers of your empirical formula to get molecular formula.