4. When 10 g of benzene (Molar mass 78 g/mol) reacts with 9 mL of concentrated HNO3 (16 M) and 9 ...
When 780 g of benzene (Molar mass = 78 g/mol) reacts with 750 mL of concentrated HNO3 (16 M) and 750 mL of H2SO4 (18 M), 1000 g of nitrobenzene (molar mass = 123 g/mol) and 250 g of an unknown compound with a high boiling point are obtained. c. Give the name of the most probable side product with the high melting point. Explain why. e. Supposing the crystalline compound (250 g) has the formula C6H4N2O4, calculate the yield...
Ester= Isoamyl acetate (banana) Molar mass of banana= 130.19 density= 0.876 g/ml 5.26 g ester product. Alcohol= isoamyl alcohol density= 0.81 g/ml Molar mass= 88.148 g/mol Quantity= 5.4 ml Carboxylic acid= acetic acid density= 1.05 g/ml Molar mass= 60.052 g/mol Quantity= 11.4 ml and 1 ml of concentrated H2SO4 (1.8 g, 0.018 mole) was also used in the procedure. Calculate theoretical yield and % yield .
For CaF2 Kap = 3.9 x 10-" (molar mass = 78 g mol-') at 25°C. How many grams of this salt can be dissolved in 100 mL water at 25°C? (A) 0.8 mg (B) 1.2 mg (C) 1.7 mg (D) 2.0 mg (E) 2.4 mg
Compound molar mass density volume mmol equ 21% sodium ethoxide in ethanol 68.05 g/mol 0.868 g/mL 20 mL ~1.3 *1-bromobutane 137.02 g/mol 1.276 g/mL 4.5 mL ~1.0 *Limiting Reagent: 1-bromobutane I need help finding the mmol of sodium ethoxide in ethanol and 1-bromobutane. Also how to find the theoretical yield of this reaction from this graph?