PLEASE HELP ME!! DIMENSIONAL ANALYSIS
CHEMISTRY
DUE IN 5 MINS!!!

12 mi/h

Explanation:

Step 1: Given data

Step 2: Convert "d" to miles

We will use the conversion factor 1 mi = 1.60934 km.

6 km × 1 mi/1.60934 km = 3.7 mi

Step 3: Convert "t" to hours

We will use the conversion factor 1 h = 60 min.

19 min × 1 h/60 min = 0.32 h

Step 4: Calculate the average speed of the runner (s)

The speed is equal to the quotient between the total distance and the time elapsed.

s = d/t

s = 3.7 mi/0.32 h = 12 mi/h

answer is d because it is trust me

explanation:

the largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. the number of potential organic compounds has been estimated at 1060—an astronomically high number. the existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. the simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. in addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. we use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. the familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. we can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. this leads to differences in geometries and in the hybridization of the carbon orbitals.