Calculate the grams of carbon in 9.32 x 10+23 molecules of benzene (C6H6).
Calculate the grams of carbon in 9.32 x 10+23 molecules of benzene (C6H6).
For the following reaction, 4.00 grams of benzene (C6H6) are allowed to react with 15.7 grams of oxygen gas.benzene (C6H6)(l) + oxygen (g)→carbon dioxide (g) + water(g)What is the maximum amount of carbon dioxide that can be formed? gramsWhat is the FORMULA for the limiting reagent?What amount of the excess reagent remains after the reaction is complete? grams
Write and balance the chemical equation for the combustion of benzene (C6H6). Also calculate how many grams of carbon dioxide will form from the reaction of 155 g of C6H6.
how many molecules in a sample of benzene, C6H6, which contains a total of 5.29 mol of carbon.
A solution is prepared by dissolving 16.2 g of benzene (C6H6) in 282 g of carbon tetrachloride (CCl4). The density of the solution is 1.55 g/mL. The molar masses of benzene and carbon tetrachloride are 78.1 g/mol and 154 g/mol respectively. What is the mass percent, mole fraction, molarity and molality of the benzene?
23. How many molecules of carbon monoxide, CO, comprise 4.82 grams?
A sample of benzene, C6H6, contains 2.5 moles. How many moles of carbon are present in this sample?
Benzene, C6H6, has a cyclic structure where the carbon atoms make a hexagon. The π electrons in the cyclic molecule can be approximated as having two-dimensional rotational motion. Calculate the diameter of this “electron ring” if it is assumed that a transition occurring at 260.0 nm corresponds to an electron going from m = 3 to m = 4.
calculate the melting point in Celsius of a solution that is 4.37% benzene (C6H6 molar mass 78.108 g/mol) by mass in carbon tetrachloride. the normal melting point of CCl4 is -22.9 Celsius and Kf= 29.9 C/m thank you!
if
3.011 x 10^23 molecules have a mass of 20.04 grams, what is the
mokar mass if this substance?
8) If 3.011 x 1023 molecules have a mass of 20.04 grams, what is the molar mass of this substance? A) 40.08 g/mol B) 10.02 g/mol C) 20.04 g/mol D) 6.658 x 10-23 g/mol E) none of the above
The nonvolatile, nonelectrolyte TNT (trinitrotoluene), C7H5N3O6 (227.1 g/mol), is soluble in benzene C6H6. How many grams of TNT are needed to generate an osmotic pressure of 8.44 atm when dissolved in 189 ml of a benzene solution at 298 K. grams TNT