Mole percent, or molar percent of a substance is the ratio of the moles of a substance in a mixture to the moles of the mixture. It represents the number of moles of a substance in a mixture as a percentage of the the total number of moles in the mixture.
Mole % = (mol substance in a mixture) / (mol mixture) * 100
MolesOne mole is 6.02 × 1023 of anything. One mole of atoms is 6.02 × 1023 atoms, one mole of rice is 6.02 × 1023 grains, one mole of shoes is 6.02 × 1023 shoes. You get the picture? One mole of molecules is 6.02 × 1023 molecules.
The base unit for the amount of a substance is an hour.
We could use a laboratory scale to measure a mole of solid material. A mole represents Avogadro's number of atoms or molecules of the solid material. Table salt, sodium chloride, for instance has a weight of 58.44277 grams per mole. The scale in the chemistry lab would be the instrument to use to measure that and to measure out a mole of most other solids.
He has a gigantic mole
6mm is a quarter inch.
No.
1.61 mole percent is equivalent to 16,100 parts per million (PPM). This conversion is based on the fact that 1 mole percent is equal to 10,000 parts per million.
No, the two concepts are quite unrelated.
No, mole percent and volume percent are not necessarily equal for a gas. Mole percent is the ratio of the moles of a gas to the total moles of all gases in a mixture, while volume percent is the ratio of the volume of a gas to the total volume of all gases in a mixture. The two can be equal only if the gases have the same molar volume at the given conditions.
.005078 g is the weight of 2 mole percent molecular iodine.
You need the balanced symbol equation for the reaction. The numbers in front of the formulae show the mole ratios. For example, in the thermal decomposition of calcium carbonate: CaCO3 --> CaO + CO2 The equation is balanced. The mole ratio between CaCO3 and CO2 is 1:1 because there is 1 mole of CaCO3 for every mole of CO2
To find the mole fraction of oxygen, first convert the percentages to fractions: 37% oxygen is 0.37 and 63% nitrogen is 0.63. Since the total mole fraction in a mixture is 1, the mole fraction of oxygen would be 0.37/(0.37 + 0.63) = 0.37/1 = 0.37. Therefore, the mole fraction of oxygen in the gas mixture is 0.37.
If i understand the question as a percent of sulfate in a compound of Cu(ii)5+(So4)5. If i can remember the formula take the total molar mass of Cu5 as 317.75g/mole and of (So4)5 480.35g/mole add them together, 317.75g/mole + 480.35g/mole this becomes your denominator. Take the mass of sulfate and place in the numerator and the outcome is the multiplied by 100. (So4)5 480.35g/mole --------------------------------------------------------- = .60187 x 100 = 60.187% Cu5 (317.75g/mole) + (So4)5 (480.35 g/mole) But this will not answer the percent if it is an amount of compound in an amount H2o you would need to separate the compound from the water then take there weights and divide them by there weights prior to separation. The above answer is theoretical based on percent of sulfate in a molecule of Cu5(So4)5.
50 mol%, because 22 g CO2 = 0.5 mole
Your question doesn't have a single answer. In Chemistry solution strength is usually calculated in moles pe liter of mole per kilogram or as a mole fraction ( moles per mole) but can also be mass per liter, mass per mass, volume per volume, percent by mass, or percent by volume.
We assume 100 grams of compound and change % to grams. Get moles. 40 grams S (1 mole S/32.07 grams) = 1.247 60 grams O (1 mole O/16.0 grams) = 3.75 Now, the smallest mole value, sulfur, is 1. Divide the oxygen mole value by the sulfur mole value. 3.75 mole O/1.247 mole S = 3.00 SO3 ------- is the empirical formula
To calculate the mass percent of benzene in the solution, we first need to convert the mole fraction of benzene to mass fraction using the molecular weights of benzene and toluene. Then, we can use the formula: Mass percent = (mass fraction of benzene / total mass of solution) x 100 Given the mole fraction of benzene as 0.40, we can use this information to determine the mass fraction and then calculate the mass percent of benzene in the solution.