All categories

3 years ago

Marijuana most abundant psychoactive chemical is delta-9- tetrahydrocannabinol or

Chemistry

2 answers:

Galina-37 [17]3 years ago

7 0

<h2>Answer:</h2>

<u>Yes the statement is</u><u> True</u>

<h2>Explanation:</h2>

THC is a chemical which stands for delta-9-tetrahydrocannibinol or Δ-9-tetrahydrocannabinol (Δ-9-THC). This chemical is a cannabinoid molecule in marijuana or cannabis that has long been known as the main psychoactive ingredient in marijuana which means that it is the substance that causes users to experience the marijuana high. It can be detected in the blood up to 20 hours after ingestion, and it's stored in the body fat and organs for three to four weeks after ingestion.

adell [148]3 years ago

4 0

Answer:

THC

Explanation:

Took the test

You might be interested in

The sun contains_____ of the mass in the solar system

alexira [117]

Answer:

99.86%

Explanation:

The sun contains 99.86% of the mass in the solar system. It is a HUGE star and it is the brightest star too. the other .14% is usually the planets and the galaxy's.

5 0

3 years ago

The currently accepted model of the atom includes an electron cloud.<br> A. True<br> B. False

natulia [17]

Answer:

True

Explanation:

4 0

3 years ago

Read 2 more answers

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

How do we know what's inside an atom?

Lemur [1.5K]

Answer:

To really 'see' inside the atom and detect fundamental particles we need to go beyond visible light and bounce particles off the smallest building blocks of the Universe to 'observe' them. This session introduces students to the scientific process—how scientific methods and theories are updated and develop as new evidence and information is collected.

Explanation:

7 0

3 years ago

How are living things put together?

Lelu [443]

D because B is silly, yes we are in groups but that is not how we are put together, Same for reproduction. That is how you're made.

7 0

3 years ago