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MOLES!!!. MOLE JEOPARDY. Category 1. Category 2. Category 3. Category 4. Category 5. 10. 10. 10. 10. 10. 20. 20. 20. 20. 20. 30. 30. 30. 30. 30. 40. 40. 40. 40. 40. 50. 50. 50. 50. 50. The number of grams in one mole of carbon…. Answer. 12 grams.
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MOLE JEOPARDY Category 1 Category 2 Category 3 Category 4 Category 5 10 10 10 10 10 20 20 20 20 20 30 30 30 30 30 40 40 40 40 40 50 50 50 50 50
12 grams From Periodic Table
2 moles of C 24 grams C x 1 mol C = 2 mol of C 12 g C
The volume in L of gas evolved from 2.0 moles of O2 at STP… Answer!
44.8 L of O2 2 mol O2 x 22.4 L O2 = 44.8 L 1 mol O2
8.43 x 10^(23) atoms 1.4 mol C x (6.02x10^(23) particles) = 8.43x10^(23) 1 mol C
The volume in L of gas evolved from 20 g of O2 at STP… Answer!
14 L of O2 20 g O2 x 1 mol O2 x 22.4 L O2 = 14 L 32 g O2 1 mol O2
22.99 grams From Periodic Table
4.32 moles Na 99 grams Na x 1 mol = 4.32 mol of Na 22.99 g
1.5 moles O2 48 grams O2 x 1 mol = 1.5 mol of O2 32 g O2
The volume in L of gas evolved from 98 grams of H2O… Answer!
121. 96 L H2O 98 g H2O x 1 mol H2O x 22.4 L H2O = 121.96 L 18 g H2O 1 mol H2O
The volume in L of gas evolved from 61 grams of CO2… Answer!
31.05 L of CO2 61 g CO2 x 1 mol x 22.4 L = 31.05 L 44 g 1 mol
40.08 grams Ca From Periodic Table!
The volume in L of gas evolved from 2.21 moles of CCl4… Answer!
49.5 L of CCl4 2.21 mol CCl4 x 22.4 L = 49.5 L CCl4 1 mol
1.81x10^(21) atoms of Ca 0.003 mol Ca x (6.02x10^(23) particles) = 1.81 x 10^(21) 1 mol Ca
The volume in L of gas evolved from 2.21 grams of CCl4… Answer!
0.322 L of CCl4 2.21 g CCl4 x 1 mol CCl4 x 22.4 L CCl4 = 0.322 L 153.8 g 1 mol
3.6x10^(23) atoms of Ca 24.2 g Ca x 1 mol Ca x (6.02x10^(23) particles = 3.6 x 10^(23) atoms 40.08 g Ca 1 mol Ca
1.008 grams 1 mol H2 x 1.008 g H2 = 1.008 g 1 mol H2
5.76 moles Br2 922 grams Br2 x 1 mol Br2 = 5.76 mol Br2 159.8 g Br2
The volume in L of gas evolved from 0.12 moles of hydrogen gas at STP… Answer!
2.688 L H2 0.12 mol H2 x 22.4 L H2 = 2.688 L 1 mol H2
The volume in L of gas evolved from 66 grams of H2 at STP… Answer!
733.3 L H2 66 g H2 x 1 mol H2 x 22.4 L H2 = 733.3 L 2.016 g H2 1 mol H2
The number of atoms found in 0.4400000000 grams of bromine gas… Answer!
1.67x10^(21) atoms of Br2 0.44 g Br2 x 1 mol Br2 x (6.02x10^(23) particles = 1.67 x 10^(21) atoms 153.8 g Br2 1 mol Br2
207.2 grams 1 mol Pb x 207.2 g Pb = 207.2 g Pb 1 mol Pb
0.169 mol Pb 35 grams Pb x 1 mol Pb = 0.169 mol of Pb 207.2 g Pb
59.59 atoms of Pb 9.9x10^(-23) mol Pb x (6.02x10^(23) particles) = 59.59 atoms of Pb 1 mol Pb
The volume in L of gas evolved from 8522 grams of Rn at STP… Answer!
859.8 L Rn 8522 g Rn x 1 mol Rnx 22.4 L Rn = 859.8 L Rn 222 g Rn 1 mol Rn