[1.72*10+23(molecules)] * [180.157 g.mol-1] / [6.02214129*10+23(molecules.mol−1]
= 51.45 g C9H8O4
0.40 moles of aspirin is 72.1 g.
1.50 moles C9H8O4 (9 moles C/1 mole C9H8O4)(6.022 X 1023/1 mole C)= 8.13 X 1024 carbon atoms===================
1 mole C4H10 = 58.1222g = 6.022 x 1023 molecules 11.7g C4H10 x 6.022 x 1023 molecules/58.1222g = 1.21 x 1023 molecules C4H10
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
1 mole HgO = 216.59g HgO = 6.022 x 1023 molecules HgO 64.0g HgO x (1mol HgO/216.59g HgO) x (6.022 x 1023 molecules HgO/mol HgO) = 1.78 x 1023 molecules HgO
0.40 moles of aspirin is 72.1 g.
1.50 moles C9H8O4 (9 moles C/1 mole C9H8O4)(6.022 X 1023/1 mole C)= 8.13 X 1024 carbon atoms===================
1 mole C4H10 = 58.1222g = 6.022 x 1023 molecules 11.7g C4H10 x 6.022 x 1023 molecules/58.1222g = 1.21 x 1023 molecules C4H10
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
1 mole HgO = 216.59g HgO = 6.022 x 1023 molecules HgO 64.0g HgO x (1mol HgO/216.59g HgO) x (6.022 x 1023 molecules HgO/mol HgO) = 1.78 x 1023 molecules HgO
Multiply the number of moles by the molecular weight.
Avogadro's number. I will show you. 18.02 grams water (1 mole H2O/18.016 grams)(6.022 X 1023/1 mole H2O) = 6.021 X 1023 atoms of water ----------------------------------------
No, there will be more molecules in the 50 grams of NaCl, because its molacular weight is lower. NaCl has a molaculair mass of 58.44 g/mol and MgCl2 has a molcular mass of 95.21 g/mol. 50 g / 58.44 g/mol = 0.86 mol NaCl 50 g / 95.21 g/mol = 0.53 mol MgCl2 The avogadro contstant states that 1 mole equals 6.02214179*1023 molecules. So you have 0.86 * 6.022*1023 = 5.18*1023 molecules of NaCl and 0.53 * 6.022*1023 = 3.19*1023 molecules of MgCl2.
divide the amount of particles given in the question by avagado's number to get the amount in moles. 3.01 x 1023 / 6.022 x 1023 which is about 0.5 moles. them multiply the amount of moles by the mass of Nitrogen to get it in grams. 0.5 x 14 = 7g
What is the weight in grams of 3.36 × 1023 molecules of copper sulfate (CuSO4)?
What is the weight in grams of 3.36 × 1023 molecules of copper sulfate (CuSO4)?
1 mole of molecules = 6.022 x 1023 molecules 1.25 moles of molecules x 6.022 x 1023 molecules/mol molecules = 7.53 x 1023 molecules