'15 mol' diaporamas de présentation

Unit 1 rates

Unit 1 rates. Go to question. 1. 2. 3. 4. 5. 6. 7. 8. An experiment was carried out at four temperatures. The table shows the times for the reaction to occur. 1. The rate of reaction is directly proportional to the temperature. b. The reaction is endothermic.

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Unit 1 rates

Unit 1 rates. Go to question. 1. 2. 3. 4. 5. 6. 7. 8. An experiment was carried out at four temperatures. The table shows the times for the reaction to occur. 1. The rate of reaction is directly proportional to the temperature. b. The reaction is endothermic.

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Solving Limiting Reactant Problems

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Empirical formula:

Empirical formula:. Hydroquinone, used as a photographic developer is 65.4% C, 5.5% H, and 29.1% O, by mass. What is the empirical formula?. If we work with 100 grams, we would have 65.4 g C, 5.5 g H and 29.1 g O. Then: Change g to moles and convert to whole numbers. 65.4 g. = 5.42 mol C.

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MOLE (mol)?

MOLE CONVERSIONS. DO NOW : Which conversion factor allows you to convert from moles to particles?. Particles ( atoms or molecules) ?. MOLE (mol)?. Which conversion factor allows you to convert from particles to moles?. 1 mol. 6.022 x 10 23 particles .

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MOLE (mol)?

MOLE CONVERSIONS. DO NOW : Which conversion factor allows you to convert from moles to particles?. Particles ( atoms or molecules) ?. MOLE (mol)?. Which conversion factor allows you to convert from particles to moles?. 1 mol. 6.022 x 10 23 particles .

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Stoichiometry

Stoichiometry. Limiting Reagents. Stoichiometry. If the quantity of each reactant is given, you must determine which one is used up first. This is the Limiting Reagent The other reactant is not completely consumed in the reaction. It is in Excess. Stoichiometry.

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???? DNA ???

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M 1 . V 1 = M 2 . V 2

NaOH = ? M. HCl = 0.10 M. M 1 . V 1 = M 2 . V 2. (0.10 M )(1.10 mL) = ( M 2 )(2.35 mL). Titration Lab Teacher Notes: Step 1) Mix 5 liters of 0.100 M HCl M 1 V 1 = M 2 V 2 (0.100 M) (5000 mL) = (12.1 M) (x mL) requires 41.3 mL of 12.1 M HCl for 5 L of 0.100 M HCl

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M 1 . V 1 = M 2 . V 2

NaOH = ? M. HCl = 0.10 M. M 1 . V 1 = M 2 . V 2. (0.10 M )(1.10 mL) = ( M 2 )(2.35 mL). Titration Lab Teacher Notes: Step 1) Mix 5 liters of 0.100 M HCl M 1 V 1 = M 2 V 2 (0.100 M) (5000 mL) = (12.1 M) (x mL) requires 41.3 mL of 12.1 M HCl for 5 L of 0.100 M HCl

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M 1 . V 1 = M 2 . V 2

NaOH = ? M. HCl = 0.10 M. M 1 . V 1 = M 2 . V 2. (0.10 M )(1.10 mL) = ( M 2 )(2.35 mL). Titration Lab Teacher Notes: Step 1) Mix 5 liters of 0.100 M HCl M 1 V 1 = M 2 V 2 (0.100 M) (5000 mL) = (12.1 M) (x mL) requires 41.3 mL of 12.1 M HCl for 5 L of 0.100 M HCl

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Molarity

Molarity. Unit of Concentration There are many units of concentration Molarity is most useful to the chemist M = moles of solute Liters of solution Liters of solution means the total volume of water and solute. If I want a liter of solution I will not use a liter of water.

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