Whatever be the substance the one gram mole of that substance would have 6.023 x 1023 atoms or molecules or ions in it. Hence to get the mole just divide the number given by 6.023 x 1023
A sample that contains 1.061023 fluorine atoms has 3,347 moles.
Okay, a mole of potassium perchlorate contains 6.02x1023 formula units of potassium perchlorate, but you're asking about individual atoms. So, let's look at the formula: KClO3. That's 1 potassium, 1 chlorine, and 3 oxygens, for a total of 5 atoms per formula unit. Now, multiple 5 by Avogadro's number above, to get 30.1x1023, which simplifies to 3.01x1024 atoms.
Are you familiar with mole concept. Well according to it 1 mole of anysubstance contains 6.023e23. First calculate the molecular mass of glucose which is (6*12)+12+(6*16)=180=y gm(say). Now find the mass of the sample glucose say x gm. Calculate the number of moles of glucose present which is found out by x/y. If its exactly 1 mole then the number of carbon atoms present are 6.023e23. Else use unitary method to find out for other values. For your mentioned number of carbon atoms the 0.0021 moles of glucose present.
We know from looking at the molecular formula that one mole of molecules of H2SO4 contains 2 moles of atoms of hydrogen, 1 mole of atoms of sulfur and 4 moles of atoms of oxygen.
1 mole of substance contains 6.02 x 1023 of its constituent particles. Accordingly, there would be 5.85 moles (of the substance) in a sample containing 3.52 x 1024 particles.
The formula mass of sodium carbonate, Na2CO3 is 2(23.0) + 12.0 + 3(16.0) = 106.0Amount of Na2CO3 = mass of sample/molar mass = 0.75/106.0 = 0.00708mol There are 0.00708 moles of Na2CO3 in a 0.75g pure sample.
Which sample contains the greatest number of atoms. A sample of Mn that contains 3.29E+24 atoms or a 5.18 mole sample of I?The sample of _____ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of I to atoms of I.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.18 moles of I to atoms of I:atoms I= 5.18 mol I6.02 x 1023 atoms I = 3.12E+24 atoms I1 mol IMultiply by atoms per mole. Moles cancel out.The sample of Mn contains 3.29E+24 atoms.Since 3.12E+24 is smaller than 3.29E+24, the sample of Mn contains the greatest number of atoms.
The mass of 4 moles of fluorine F atoms is 151,98 g (because fluorine is a diatomic element).
Which sample contains the greatest number of atoms. A sample of Al that contains 8.18E+23 atoms or a 5.16 mole sample of S?The sample of______ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of S to atoms of S.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.16 moles of S to atoms of S:atoms S=5.16 mol S6.02 x 1023 atoms S= 3.11E+24 atoms S1 mol SMultiply by atoms per mole. Moles cancel out.The sample of Al contains 8.18E+23 atoms.Since 3.11E+24 is larger than 8.18E+23, the sample of S contains the greatest number of atoms.
One: The formula for magnesium fluoride is MgF2. Since each mole of fluorine molecules, which have the formula F2, contains two moles of fluorine atoms, one mole of each is the right ratio.
1.50 x 10 to the 23 atoms of fluorine is equal to 0,249 moles.
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
each mole has avogadro's number of atoms in it, which is 6.02 x 10 to the 23rd power. so, 4 x 6.02 x 10 23 power of atoms
The molecular mass of fluorine gas, F2 is 2(19.0) = 38.0Amount of F2 = mass of sample/molar mass = 9.5/38.0 = 0.25mol There are 0.25 moles of fluorine in a 9.5g pure sample.
0,25 moles F2
Approx 7.925*10^23 atoms.
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The gram atomic mass of phosphorus is 30.9738, and by definition, a mole of such atoms contains Avogadro's Number of atoms. Therefore, 100 g of phosphorus contains 100/30.9738 or 3.23 moles, to the justified number of significant digits.