The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol (c2h5oh) present. calculate the number of grams of alcohol present in 1.00 l of 48-proof gin. the density of ethanol is 0.798 g/ml.

m C2H5OH = 191.52 g

Explanation:

∴ %v/v C2H5OH = ( v C2H5OH / v sln)×100

⇒ 48 proof = (2) (%v/v C2H5OH)

⇒ 48/2 = %v/v C2H5OH

⇒ 24 = %v/v C2H5OH

⇒ 24/100 = v C2H5OH / v sln = 0.24

∴ v sln (gin) = 1.00 L

⇒ v C2H5OH = ( 0.24 )( 1.00 L )

⇒ v C2H5OH = 0.24 L = 240 mL

⇒ m C2H5OH = (240 mL)(0.798 g/mL)

⇒ m C2H5OH = 191.52 g

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.

2b2t

explanation:

2b2t