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How many grams of N combine with 7.05x1022 O atoms in N2O5?
The formula N2O5 shows that there are 2/5 as many nitrogen atoms as oxygen atoms in the compound. Therefore, the number of nitrogen atoms required is (2/5)(7.05 X 1022) or 2.82 X 1022 atoms. The gram atomic mass of nitrogen is 14.0067 and, by definition, consists of Avogadro's Number of atoms. Therefore, the mass of nitrogen required to react with the specified amount of oxygen to produce the specified compound is 14.0067 [(2.82 X 1022)/(6.022 X 1023] or 0.656 grams of nitrogen, to the justified number of significant digits.
How many atoms are in 0.7 grams of Helium?
.7 g= .17 mol = 1.05 x 10^23 atoms
What is the approx mass of one mole of helium atoms?
4 grams. Approximately, mass (in grams) of 1 mole of any chemical element is just the atomic mass of this element (4 - for He, 12 - for C, and so on). It follows from the definition of one mole - number of atoms in 12 g of C.
How many individual atom make up one unit of aluminium chloride?
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023. AlCl3= 133.5 grams10.5 grams Cu / (133.5 grams) × (6.02 × 1023 atoms) = 4.73 × 1022 atoms
How many atoms in 3 moles helium?
For this you need the atomic (molecular) mass of H2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.3.00 moles H2 × 2.02 = 6.06 grams H2
What is the mass in grams of 3.40x10 22 atoms He?
3.40 X 1022 atoms helium (1 mole He/6.022 X 1023)(4.003 grams/1 mole He) = 0.226 grams ===========
How many atoms of gold are in 5 grams of gold?
Atomic mass of Ag: 107.9 grams5.00 grams × (6.02 × 1023 atoms) / (107.9 grams) = 2.79 × 1022 atoms Ag
How many atoms are in 1.9 g Carbon?
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
How many grams do 4.4 x 1022 atoms of potassium weigh. answer in units of grams?
The mass is 2.86 grams but the weight will be 0.028 Newtons.
What is the mass of 5.0g of Fe how many atoms are present?
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.5.0 grams Fe / (55.9 grams) × (6.02 × 1023 atoms) = 5.38 × 1022 atoms
How many atoms of sulfur are in 3.86 g?
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.3.86 grams S / (32.1 grams) × (6.02 × 1023 atoms) = 7.24 × 1022 atoms
What is the mass in grams of 3.50mol LiBr?
step 1: calculate the molar mass of one molecule LiBr by adding together their amu (atomic mass units) step 2: 3.50 mol * ( molecular mass LiBr / 1 mol LiBr ) = grams LiBr amu on your periodic table is grams when you have one mol of that substance
What is the mass of 4.771E plus 22 atoms of neodymium?
4.771 X 1022 atoms neodymium (1 mole Nd/6.022 X 1023)(144.2 grams/1 mole Nd) = 11.42 grams of neodymium --------------------------------------
What is the mass of one mole atoms?
The mass depends on the element whose atoms are concerned. For eg, 1 mol Helium atoms have mass 4 g 1 mol Neon atoms have mass 20 g.
How many Helium-atoms and Carbon-atoms do you have if you have He-atoms with mass of 40amu and C-atoms with mass of 120amu?
Mass of 1 Helium atom is 4 amu and mass of 1 Carbon atom is 12 amu. So there are 10 helium atoms and 10 carbon atoms
How many atoms are present in 3.14g of Cu?
3.14 g Cu = # atomsTake the known mass of copper multiply it by Avogadro number and divided by the atomic weight.Atomic weight of copper:63.5 g3.14 g Cu (6.02 × 1023 atoms) / (63.5 grams) = 2.98 × 1022 atoms of Copper
How many grams of N combine with 7.05x1022 O atoms in N2O5?
The formula N2O5 shows that there are 2/5 as many nitrogen atoms as oxygen atoms in the compound. Therefore, the number of nitrogen atoms required is (2/5)(7.05 X 1022) or 2.82 X 1022 atoms. The gram atomic mass of nitrogen is 14.0067 and, by definition, consists of Avogadro's Number of atoms. Therefore, the mass of nitrogen required to react with the specified amount of oxygen to produce the specified compound is 14.0067 [(2.82 X 1022)/(6.022 X 1023] or 0.656 grams of nitrogen, to the justified number of significant digits.
