Find the atomic or molecular weight of each and multiply it by the number of moles.
atomic weight of carbon is 12 g/mole.
atomic weight of chlorine is 35.45 g/mole
molecular weight of fructose is 180.16 g/mole
So just find the total grams.
10 moles Carbon * 12g/mole Carbon = 120 grams carbon
3 moles Chlorine * 35.45g/mole Chlorine = 106.35 grams Chlorine
1 mole Fructose * 180.16 g/mole Fructose = 180.16 grams Fructose
So the answer is 1 mole of fructose.
The mass in grams of one mole H2SO4 is: 98.079 g/mol (molar mass)The mass in grams of one molecule H2SO4 is (molecule mass):98.079 (g/mol) / 6.02214*10+23(molecules/mol) = 16.286*10-23 g/molecule
#moles = mass/molar mass mass = #moles*molar mass mass = .10 moles*(atomic weight of na+atomic weight of N+3(atomic weight of oxygen)
You calculate it by using Avogadro's number: 6.022 x10 23 which is one mole of substance. 1.00 x 1024 divided by 6.022 x 1023 = 10 divided by 6.022 = 1.661 mol H2O Use the molar mass of water (18.02 g mol-1) to get to the mass in grams. 18.02 g x 1.661 mol = 29.93 g H2O
From my understanding as there is only one mole of zinc which means the mass is 65.4 . Then to get the number of particles the answer is simply 1(the number of moles) x 6.02 x 10 ^ 23. is this correct this is not the answer just a further question ...
The mass defect represents the mass converted to binding energy
The mass in grams of one mole H2SO4 is: 98.079 g/mol (molar mass)The mass in grams of one molecule H2SO4 is (molecule mass):98.079 (g/mol) / 6.02214*10+23(molecules/mol) = 16.286*10-23 g/molecule
Since magnesium and helium do not form elemental molecules, a mole of these elements is the same as a gram atomic mass, which is 24.305 for magnesium and 4.00260 for helium. The gram molecular mass for sucrose is 342.30. Therefore, the mass of: A. 3 moles of magnesium is 72.915 grams; B 1 mole of sucrose (C12H22O11) is 342.30 grams; and C. 10 moles of helium is 40.0260 grams. The largest of these is obviously the single mole of sucrose.
Each mole of particles have 6.02 x 10^23 particles. (3.6 x 10^20) / (6.02 x 10^23) = 0.000598 mol of Silicon Ar of Si (Silicon) = 28.1g/mol mass = number of moles x Ar mass = 0.000598 mol x 28.1g/mol = 0.0168g of silicon
Mass (g) = Number of moles (mol) x Molar mass (g/mol) Here are the calculations for each sample: 7 moles of aluminum (Al): The molar mass of Al is approximately 26.98 g/mol. Mass = 7 mol x 26.98 g/mol = 188.86 g 1.400e-15 teramoles (Tmol) of carbon dioxide (CO2): The molar mass of CO2 is approximately 44.01 g/mol. Mass = 1.400e-15 mol x 44.01 g/mol = 6.16e-14 g 7.7e12 picomoles (pmol) of sodium nitrate (NaNO3): The molar mass of NaNO3 is approximately 85 g/mol. Mass = 7.7e12 mol x 85 g/mol = 6.59e-10 g 2.73e-4 kilomoles (kmol) of C2H6O: The molar mass of C2H6O (ethanol) is approximately 46.07 g/mol. Mass = 2.73e-4 mol x 46.07 g/mol = 0.1256 g 1.02e-9 megamoles (Mmol) of uranium (U): The molar mass of uranium is approximately 238.03 g/mol. Mass = 1.02e-9 mol x 238.03 g/mol = 2.43e-7 g 4.01 millimoles (mmol) of hydrobromic acid (HBr): The molar mass of HBr is approximately 80.91 g/mol. Mass = 4.01e-3 mol x 80.91 g/mol = 0.3247 g So, the mass of each sample is as follows: 188.86 grams of aluminum 6.16e-14 grams of carbon dioxide 6.59e-10 grams of sodium nitrate 0.1256 grams of C2H6O 2.43e-7 grams of uranium 0.3247 grams of hydrobromic acid.
6.022*10**23 atoms / mol = avagadro's constant 63.546 g / mol = atomic weight of copper 1 atom / 6.022*10**23 atoms/mol * 63.546 g/mol = 1.05523082*10**-22g 1 g / 63.546 g/mol * 6.022*10**23 atoms/mol = 9.476599629*10**21 atoms
(g) -> moles = x(g) * (1 mol/molar mass)Moles -> atoms = x(mol) * (6.022*10^23/1 mol)
Balanced equation: LiOH + HBr ---> LiBr + H₂O Here, we aim to convert the mass of LiOH to mass of LiBr. In this formula, the product (LiBr) takes x, and the reactant (LiOH) takes y. Here's how it goes. (? = coefficient in the balanced equation) mass of x = (mole of y) * (? mol x / ? mol y) * (molar mass of x) mass of LiBr = (10 g / 23.95 g/mol) * (1 mol LiBr / 1 mol LiOH) * (86.85 g/mol LiBr) mass of LiBr = 36.3 g (Answer)
To find the mass of carbon as a percent of the molecule, you have to first know the mass of the whole molecule. This can be done with the atomic weights of the elements and multiplying them by the number of atoms in a single molecule of hexachlorophene (the given compound).Carbon: 12.0 g/mol × 13 = 156Hydrogen: 1.0 g/mol × 6 = 6.0Chlorine: 35.5 g/mol × 6 = 213Oxygen: 16.0 g/mol × 2 = 32.0C13H6Cl6O2 = 407 g/mol(Mass of carbon ÷ total mass) × 100 = % carbon by mass(156 ÷ 407) × 100 = 38.3% carbon
Multiply by molar mass of NaCl:Molar mass of Na = 22.989 g/mol NaMolar mass of Cl = 35.453 g/mol ClTotal molar mass = 58.442 g/mol NaClThe answer:58.4398 g/mol * 0.5 mol = 29.221 g= 29 g (rounded according to the least significant number)
#moles = mass/molar mass mass = #moles*molar mass mass = .10 moles*(atomic weight of na+atomic weight of N+3(atomic weight of oxygen)
You calculate it by using Avogadro's number: 6.022 x10 23 which is one mole of substance. 1.00 x 1024 divided by 6.022 x 1023 = 10 divided by 6.022 = 1.661 mol H2O Use the molar mass of water (18.02 g mol-1) to get to the mass in grams. 18.02 g x 1.661 mol = 29.93 g H2O
Formula for table salt = NaCl 1.00 g = 1.00 / 58.45 mol = 1.71 x 10-2 mol Mole Na required = 1.71 x 10-2 mol Mass of Na = 1.71 x 10-2 mol x 23.0 g /mol = 0.393 g = 393 mg