Answer:
For much of the past century, scientists studying drugs and drug use labored in the shadows of powerful myths and misconceptions about the nature of addiction. When scientists began to study addictive behavior in the 1930s, people with an addiction were thought to be morally flawed and lacking in willpower. Those views shaped society’s responses to drug use, treating it as a moral failing rather than a health problem, which led to an emphasis on punishment rather than prevention and treatment.
Today, thanks to science, our views and our responses to addiction and the broader spectrum of substance use disorders have changed dramatically. Groundbreaking discoveries about the brain have revolutionized our understanding of compulsive drug use, enabling us to respond effectively to the problem.
As a result of scientific research, we know that addiction is a medical disorder that affects the brain and changes behavior. We have identified many of the biological and environmental risk factors and are beginning to search for the genetic variations that contribute to the development and progression of the disorder. Scientists use this knowledge to develop effective prevention and treatment approaches that reduce the toll drug use takes on individuals, families, and communities.
Despite these advances, we still do not fully understand why some people develop an addiction to drugs or how drugs change the brain to foster compulsive drug use. This booklet aims to fill that knowledge gap by providing scientific information about the disorder of drug addiction, including the many harmful consequences of drug use and the basic approaches that have been developed to prevent and treat substance use disorders.
At the National Institute on Drug Abuse (NIDA), we believe that increased understanding of the basics of addiction will empower people to make informed choices in their own lives, adopt science-based policies and programs that reduce drug use and addiction in their communities, and support scientific research that improves the Nation’s well-being.
Answer:
Final volume of the gas is 4.837mL
Explanation:
Initial volume (V1) = 350mL = 0.35L
Initial temperature (T1) = 0°C = (0 + 273.15)k = 273.15k
Initial pressure (P1) = 1.0atm
Final volume (V2) = ?
Final temperature (T2) = 10°C = (10 + 273.15)k = 283.15K
Final pressure (P2) = 75atm
To solve this question, we'll have to use combined gas equation which is the combination of all gas law I.e Charle's laws, Boyle's law, Pressure law etc.
According to combined gas equation,
(P1 × V1) / T1 = (P2 × V2) / T2
Make V2 the subject of formula,
V2 = (P1 × V1 × T2) / (P2 × T1)
V2 = (1.0 × 0.35 × 283.15) / (75 × 273.15)
V2 = 99.1025 / 20,486.25
V2 = 0.004837L
V2 = 4.837mL
The final volume of the gas is 4.837mL
The right subject for this question is physics.
To calculate the work you use the formula:
Work = force * displacement
Work = 2500 pounds * 30 feet = 75,000 pounds - feet
To calculate the power you use the formula:
power = work / time
Power = 75,000 pound - feet / 30 seconds = 2300 pound-feet / second.
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
