Comments

• 2 KNO3 → 2 KNO2 + O2 (56.0 g KNO3) / (101.1032 g KNO3/mol) x (1 mol O2 / 2 mol KNO3) = 0.27694 mol O2 V = nRT / P = (0.27694 mol) x (0.08205746 L atm/K mol) x (25 + 273) K / (1.19 atm) = 5.69 L O2

  in Potassium nitrate decomposes to potassium nitrite and oxygen: 2KNO3(s)→Δ2KNO2(s)+O2(g)? Comment by roger-the-mole June 2021
• 2 KNO3 → 2 KNO2 + O2 (56.0 g KNO3) / (101.1032 g KNO3/mol) x (1 mol O2 / 2 mol KNO3) = 0.27694 mol O2 V = nRT / P = (0.27694 mol) x (0.08205746 L atm/K mol) x (25 + 273) K / (1.19 atm) = 5.69 L O2

  in Potassium nitrate decomposes to potassium nitrite and oxygen: 2KNO3(s)→Δ2KNO2(s)+O2(g)? Comment by roger-the-mole June 2021
• 1. 90 kPa = 675 mm Hg (5000 mL) x (720 / 675) x (44 + 273) / (27 + 273) = 5640 mL 2. 2 KClO3 → 2 KCl + 3 O2 (0.250 L O2) / (22.4 L/mol) x (2 mol KClO3 / 3 mol O2) x (122.5498 g KClO3/mol) / (1.20 g) = 0.760 = 76.0% KClO3 3. (1600 …

  in ¿Problema de Gases Química! Ayuda! 100 pts!? Comment by roger-the-mole June 2021
• x^3+7x^2-30x Factor out the common factor: x (x^2 + 7x - 30) Factor what remains in parentheses: x (x - 3) (x + 10)

  in How do you factor x^3+7x^2-30x? Comment by roger-the-mole June 2021
• Make the simplifying substitution: z = (3a-2b): 3z^2 + z - 14 Factor: (z - 2) (3z + 7) Undo the substitution: (3a - 2b - 2) (3a - 2b + 7)

  in How do you factor , 3(3a-2b)^2 +3a-2b-14? Comment by roger-the-mole June 2021
• To answer just the questions as stated: (2 mol O/mol) x (15.9994 g O/mol) x (100/24.38) = 131.3 g/mol (2 atoms O) x (15.9994 g O/mol) x (54.95/24.38) / (12.0108 g C/mol) = 6 atoms C The complete solution goes like this: 100% - 54.95% C - 2…

  in Two part Chemistry problem? Comment by roger-the-mole May 2021
• To answer just the questions as stated: (2 mol O/mol) x (15.9994 g O/mol) x (100/24.38) = 131.3 g/mol (2 atoms O) x (15.9994 g O/mol) x (54.95/24.38) / (12.0108 g C/mol) = 6 atoms C The complete solution goes like this: 100% - 54.95% C - 2…

  in Two part Chemistry problem? Comment by roger-the-mole May 2021
• Keep them in tightly sealed containers to keep them away from the humidity in the air.

  in how do we preserve deliquescent substances? Comment by roger-the-mole May 2021
• CaCN2 + 3 H2O → CaCO3 + 2 NH3 (109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2 (82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O 1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there…

  in CaCN2 + 3H2O --> CaCO3 + 2NH3? Comment by roger-the-mole May 2021
• CaCN2 + 3 H2O → CaCO3 + 2 NH3 (109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2 (82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O 1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there…

  in CaCN2 + 3H2O --> CaCO3 + 2NH3? Comment by roger-the-mole May 2021
• Keep them in tightly sealed containers to keep them away from the humidity in the air.

  in how do we preserve deliquescent substances? Comment by roger-the-mole May 2021
• Keep them in tightly sealed containers to keep them away from the humidity in the air.

  in how do we preserve deliquescent substances? Comment by roger-the-mole May 2021
• CaCN2 + 3 H2O → CaCO3 + 2 NH3 (109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2 (82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O 1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there…

  in CaCN2 + 3H2O --> CaCO3 + 2NH3? Comment by roger-the-mole May 2021
• 7 x 4 x 6 mm = 0.7 x 0.4 x 0.6 cm 3 x 2 x 4 mm = 0.3 x 0.2 x 0.4 cm

  in How do you convert mm to cm? Comment by roger-the-mole May 2021
• That doesn't factor in rational numbers. You can prove it by examining the discriminant.

  in how do you factor.........? Comment by roger-the-mole May 2021
• That doesn't factor in rational numbers. You can prove it by examining the discriminant.

  in how do you factor.........? Comment by roger-the-mole May 2021
• That doesn't factor in rational numbers. You can prove it by examining the discriminant.

  in how do you factor.........? Comment by roger-the-mole May 2021
• 7 x 4 x 6 mm = 0.7 x 0.4 x 0.6 cm 3 x 2 x 4 mm = 0.3 x 0.2 x 0.4 cm

  in How do you convert mm to cm? Comment by roger-the-mole May 2021
• CaCN2 + 3 H2O → CaCO3 + 2 NH3 (109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2 (82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O 1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there…

  in CaCN2 + 3H2O --> CaCO3 + 2NH3? Comment by roger-the-mole May 2021
• CaCN2 + 3 H2O → CaCO3 + 2 NH3 (109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2 (82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O 1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there…

  in CaCN2 + 3H2O --> CaCO3 + 2NH3? Comment by roger-the-mole May 2021
• Keep them in tightly sealed containers to keep them away from the humidity in the air.

  in how do we preserve deliquescent substances? Comment by roger-the-mole May 2021
• Keep them in tightly sealed containers to keep them away from the humidity in the air.

  in how do we preserve deliquescent substances? Comment by roger-the-mole May 2021
• To answer just the questions as stated: (2 mol O/mol) x (15.9994 g O/mol) x (100/24.38) = 131.3 g/mol (2 atoms O) x (15.9994 g O/mol) x (54.95/24.38) / (12.0108 g C/mol) = 6 atoms C The complete solution goes like this: 100% - 54.95% C - 2…

  in Two part Chemistry problem? Comment by roger-the-mole May 2021
• To answer just the questions as stated: (2 mol O/mol) x (15.9994 g O/mol) x (100/24.38) = 131.3 g/mol (2 atoms O) x (15.9994 g O/mol) x (54.95/24.38) / (12.0108 g C/mol) = 6 atoms C The complete solution goes like this: 100% - 54.95% C - 2…

  in Two part Chemistry problem? Comment by roger-the-mole May 2021
• Make the simplifying substitution: z = (3a-2b): 3z^2 + z - 14 Factor: (z - 2) (3z + 7) Undo the substitution: (3a - 2b - 2) (3a - 2b + 7)

  in How do you factor , 3(3a-2b)^2 +3a-2b-14? Comment by roger-the-mole May 2021
• x^3+7x^2-30x Factor out the common factor: x (x^2 + 7x - 30) Factor what remains in parentheses: x (x - 3) (x + 10)

  in How do you factor x^3+7x^2-30x? Comment by roger-the-mole May 2021
• 2 KNO3 → 2 KNO2 + O2 (56.0 g KNO3) / (101.1032 g KNO3/mol) x (1 mol O2 / 2 mol KNO3) = 0.27694 mol O2 V = nRT / P = (0.27694 mol) x (0.08205746 L atm/K mol) x (25 + 273) K / (1.19 atm) = 5.69 L O2

  in Potassium nitrate decomposes to potassium nitrite and oxygen: 2KNO3(s)→Δ2KNO2(s)+O2(g)? Comment by roger-the-mole May 2021