Why is it best to use a white card, bowl or spoon when
testing with the universal
indicator?

Since the gas in your graduated cylinder is a mixture of butane and water vapor, you must determine the partial pressure of the butane, pbutane, alone. to do this, consult a reference and record the partial pressure of the water vapor, pwater, at the temperature you recorded. use the following formula to compute the partial pressure of the butane. pbutane = atmosphere - pwater use the following combined gas law formula and compute the volume that the butane sample will occupy at stp. (hint: convert both temperatures to kelvin.) pbutane x voriginal = pstandard x vfinal troom tstandard use the following ratio and proportion formula to determine the mass of butane needed to occupy a volume of 22.4 l at stp. grams of butane you used “x” grams of butane ml of butane corrected to stp = 22,400 ml compute the theoretical molar mass of butane based on its formula and the atomic masses on the periodic table. compare your experimental results from #3 to the theoretical value of #4, computing a percent error of your findings using this formula: % error = measured value - accepted value x 100 accepted value use the following ratio and proportion formula to determine the mass of butane needed to occupy a volume of 22.4 l at stp. need asap

Part 1: include important facts found through your research. part 2: include your visual display. include your summary of “the chemistry of water” from the national science foundation website. include your experiment. part 3: include responses to the reflection questions.

Using this reversible reaction, answer the questions below: n2o4 2no2 (colorless) (reddish-brown) -as the temperature increased, what happened to the n2o4 concentration? -was the formation of reactants or products favored by the addition of heat? -which reaction is exothermic? right to left or left to right? -if the change of enthalpy of this reaction when proceeding left to right is 14 kcal, which chemical equation is correct? n2o4 2no2 + 14 kcal n2o4 2no2, hr = +14 kcal n2o4 + 14 kcal 2no2 n2o4 2no2, hr = -14 kcal