4.5165x10^10 ions x 1 mole ions/6.02x10^23 ions = 7.5025 x 10^-5 moles
The half-life of bismuth-209 is approximately 1.9 x 1019 years.
To calculate the number of molecules of allicin in 2.60 mg, you first need to convert the mass to moles using the molar mass of allicin. Then, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
The number of Avogadro states that 1 mole equals 6.02214179*10^23 mole. So if you have 1*10^19 HCl molecules, you have: 1*10^19 / 6.02214179*10^23 = 1.660538783*10^-5 mole of HCl. Which equals 0.0166053878 milimole.
There are about 6.24 x 1018 electrons in a coulomb. If we take 1.63 times that we get 1.02 x 1019 electrons. To "micro" that, we have to multiply it by 10-6, and that takes us to 1.02 x 1013 electrons. About.
The formula for the most common form of ammonium phosphate is (NH4)3PO4.3 H2O, and its gram formula mass is 203.13. The formula shows that there are 3 ammonium ions in each formula unit. 10.7g/203.13 is 5.27 X 10-2 formula units. Therefore, the number of ammonium ions present in 10.7g of this ammonium phosphate is 3 X 5.27 X 10-2 X Avogadro's Number or 9.52 X 1019 ammonium ions, to the justified number of significant digits.
They are 1 and 9.
997, 1009, 1013, 1019 . . .
1019, 1021
Three 1009 1013 1019
1019 = 1,019
The positive integer factors of 1019 are: 1, 1019
As in the question it is 1019
It is simply 1019/1 as an improper fraction
0.0206
1019 1021 1031 1033 and 1039. In fact those are the first 5 four digit primes.
Indeed, the number 1019 is prime - the number has no factors other than itself and 1.
1019