If the caffeine concentration in a particular brand of soda is 2.85 mg/oz, drinking how many cans of soda would be lethal? assume that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can.

2,375 cans

Explanation:

The strategy here is to use the information given to calculate the lethal dosage contained in the number of cans we will compute.

We know the lethal dosage is

Ld = 10.0 g caffeine

and we also know that the oncentration of caffeine is:

2.85 mg/ oz

So our problem simplifies to calculate how many oz will contain the lethal dose, and then given the ounces per can determine how many cans are required.

First convert the lethal dose in grams to mg:

Ld =( 10 g x 1000 mg ) = 10,000 mg caffeine

10,000 mg x ( 1 Oz / 2.85 mg ) = 28,500 oz

28500 oz x ( 1 can/12 oz ) = 2,375 cans

We could also have calculated it in one step  using conversion factors:

Number of cans = 10000 mg x 1 oz/ 2.85 mg x 1 can / oz = 2,375 cans

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

2b2t

explanation:

2b2t