One mole takes 22.4l at STP.So 3.6l has 0.16mol
To find the number of oxygen molecules in the balloon, you can first calculate the number of moles of O2 using its molar mass (32 g/mol). Then, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. In this case, the balloon contains approximately 1.15 x 10^23 oxygen molecules.
The number of molecules is 0,90332112855.10e23.
To blow up a 1.2 balloon, it requires 0.18 moles of exhaled air (3 breaths x 0.060 moles/breath). For a 3.0 balloon, which is 2.5 times larger, it would require 0.45 moles of air (0.18 moles x 2.5).
Using the ideal gas law, we can calculate the number of moles in the 75.0-L balloon by setting up a proportion using the volume and moles relationship between the two balloons. By cross-multiplying, we find that the 75.0-L balloon will hold approximately 4.85 moles of helium.
There are 4.17 moles of H2O present in 75.0g of H2O.
To find the number of oxygen molecules in the balloon, you can first calculate the number of moles of O2 using its molar mass (32 g/mol). Then, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. In this case, the balloon contains approximately 1.15 x 10^23 oxygen molecules.
The number of molecules is 0,90332112855.10e23.
Since each N2O molecule contains 2 nitrogen atoms, the number of moles of N2O molecules would be half of the moles of nitrogen atoms. Therefore, in this case, there would be 2.615 moles of N2O molecules present in the sample.
To blow up a 1.2 balloon, it requires 0.18 moles of exhaled air (3 breaths x 0.060 moles/breath). For a 3.0 balloon, which is 2.5 times larger, it would require 0.45 moles of air (0.18 moles x 2.5).
6,00 g of water contains how 0,33 moles.
The answer is 8,33 moles.
The answer is 0,615 moles.
Using the ideal gas law, we can calculate the number of moles in the 75.0-L balloon by setting up a proportion using the volume and moles relationship between the two balloons. By cross-multiplying, we find that the 75.0-L balloon will hold approximately 4.85 moles of helium.
Using the ideal gas law, at STP (standard temperature and pressure), 1 mole of gas occupies 22.4 liters. Therefore, a balloon with 560 liters at STP would contain 25 moles of gas (560 liters / 22.4 liters/mole).
There are 4.17 moles of H2O present in 75.0g of H2O.
The answer is 14,93 moles.
1 mol of hydrogen ion () corresponds to 1 equivalent. Therefore, 2.75 moles of in a solution would contain 2.75 equivalents.