Asubatomic particle possessing no electrical charge is called a(n a.) ion b.) neutron c.) atom d.) electron

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

Which of the following are sources of revenue for media companies? a. direct sales to producers b. advertising and subscriptions c. online purchase d. capital investments

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes