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Question 2 0/1 pts In doing a stoichiometry problem, the only one of the "three general...
Question 13 (2 points) In a particular stoichiometry problem, you are asked to find the number of moles of a reactant that will form 3 moles of one of the products in the reaction. Which of the following conversion factors would you use for this problem? (2 pts) The molar mass of the reactant Avogadro's number The mole ratio from the balanced equation. The molar volume
1 pts Question 7 2 H NaOH EOH benzaldehyde d 1.044 acetone d 0.791 When doing part B of the reaction for lab, how many grams of acetone (#5) would you need if you are trying to make a maximum theoretical yield of 15 grams of dibenzalacetone (#6)? Acetone: density 0.791 g/mL and molar mass 58.08 grams/mole 1044 g/mL and molar mass 106.12 grams/mole Benzaldehyde: density Dibenzalacetone: molar mass 234.29 grams/mole
Question 8 1 pts NAOH BOHO acetone d 0.791 benzaldehyde d 1.044 When doing part of the reaction for lab, how many grams of benzaldehyde (#4) would you need if you are trying to make a maximum theoretical yield of 10 grams of dibenzalacetone (#6)? Acetone: density = 0.791 g/mL and molar mass = 58.08 grams/mole Benzaldehyde: density - 1.044 g/mL and molar mass = 106.12 grams/mole Dibenzalacetone: molar mass = 234.29 grams/mole
can you respond to the 3 questions in the second photo
Mole Ratios and Reaction Stoichiometry Reaction A: Sodium Bicarbonate and Hydrochloric Acid Experimental Data (a) Mass of evaporating dish + watch glass (6) Mass of evaporating dish + watch glass + sodium bicarbonate (c) Mass of sodium bicarbonate used (d) Mass of evaporating dish + watch glass + sodium chloride (e) Mass of sodium chloride collected (experimental yield) 164.79 165.19 10. 49 165.19 164.7 9 Data Analysis 1) Use...
POGIL-Stoichiometry How do chemists use balanced chemical equations? got bit D 23 mosquitoes? 10 He got Mol-aria Why? Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. This is called Stoichiometry- (stoi-key-ah-meh-tree) Another way of looking at it is using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation. A mole ratio...
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Green Limiting Reagent IN ne Write the balanced chemical equation for the reaction and complete the table below. Be sure use the correct number of significant figures. Remember that leading zeros are not considered nificant Reaction = Initial moles Change moles End moles End moles numerical + What does x equal? Calculate the molar masses for...
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15. The Basic Problem In this section we will practice calculating the amounts of products produced in a chemical reaction using the three-step method. This method can be shortened later by combining the steps. 1 mole Step 1. Convert to moles: multiply grams by Tarla mass is grams Step 2. Use the chemical equation to relate the products to the reactants. Multiply the answer from Step 1 by the confident of the...
4 NH₃ + 3 0₂ 2 N2 + 6 H₂O 31 0-26 0 + 36 5. If 38.7 grams of NH3 react with excess oxygen, how many grams of nitrogen are produce (based on the balanced equation in question 4) (6pts) Smoler NH₂ moss/molor mass = 38 2117=2.276 moles moles af No = 1/2*mole sof NH3 = 1/2 * 2.276= 1.138 molos 1,138*28= 31,8649 6. If only 17.9 grams of nitrogen are produced, what is the percent yield (based on...
15. The Basic Problem In this section we will practice calculating the amounts of products produ using the three-step method. This method can be shortened later by como i mole Step 1. Convert to moles: multiply grams by formula mass in gram cts produced in a chemical reaction later by combining the steps. Step 2. Use the chemical equation to relate the products to the reactants. tiply the answer from Sten 1 The coefficient of the product you are calculating...
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4. Approximate the number of moles of hydrogen peroxide at the equivalence point in the graph in the introduction, supposing a 3.00% m/m solution. Thus the densities will be- Trial Mass(g) 0.448 0.450 3 Density(g/ml) 0.448 g/ 0.400 ml = 1.12 g/ml 0.450 g/ 0.400 ml = 1.125 g/ml 0.437 g/ 0.400 ml = 1.0925 g/m 0.442 g/ 0.400 ml = 1.105 g/ml...