Recreational drug users often obtain the drug as a white powder with unknown purity. Some users snort the power, whereas others make a solution and take it orally or inject it. To achieve a ketamine high, users usually need to take between 0.1 mg/kg and 1 mg/kg of ketamine. However, taking 2 mg/kg can lead the user to experience a K-hole or to enter an unconscious state.

If the user dissolves exactly 1 g of ketamine in 1/4 cup of water, what is the concentration of ketamine in milligrams per milliliter (mg/mL)? 1 cup=236.5 mL
What volume, in milliliters, of the recreational ketamine in the previous question would a 59.0 kg user need to take to experience a high at 0.600 mg/kg?
How many milliliters would that same user need to take to become unconscious at 2.00 mg/kg?

Answers

Answer 1
If a user dissolves exactly 1 g of ketamine in 1/4 cup of water, the resulting concentration is 16.91 mg/mL. A 59.0 kg user would have to consume 2.09 mL of this solution to experience a high of 0.600 mg/kg. The same user would have to consume 6.98 mL of this solution to become unconscious at 2.00 mg/kg.

Part 1

A user dissolves exactly 1 g of ketamine in 1/4 cup of water.

Given 1 cup = 236.5 mL, the volume of 1/4 cup of water is:

1/4 cup × (236.5 mL/1 cup) = 59.13 mL

1 g of ketamine is dissolved in 59.13 mL of water. The concentration of ketamine, in mg/mL, is:

C = 1000 mg / 59.13 mL = 16.91 mg/mL

Part 2

The volume of this solution that a 59.0 kg user (body mass) would have to consume to experience a high at 0.600 mg/Kg is:

[tex]59.0 kgBM \times \frac{0.600mgKetamine}{1kgBM} \times \frac{1mL Soluiton}{16.91mgKetamine} = 2.09 mL[/tex]

Part 3

The volume of this solution that a 59.0 kg user (body mass) would have to consume to become unconscious at 2.00 mg/kg is:

[tex]59.0 kgBM \times \frac{2.00mgKetamine}{1kgBM} \times \frac{1mL Soluiton}{16.91mgKetamine} = 6.98 mL[/tex]

If a user dissolves exactly 1 g of ketamine in 1/4 cup of water, the resulting concentration is 16.91 mg/mL. A 59.0 kg user would have to consume 2.09 mL of this solution to experience a high of 0.600 mg/kg. The same user would have to consume 6.98 mL of this solution to become unconscious at 2.00 mg/kg.

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Related Questions

Methane (CH4) is used in laboratory burner. When 1 mole of methane burns at constant pressure, it produces 804 kJ of heat and does 3 kJ of work.Methane (CH4) is used in laboratory burner. When 1 mole of methane burns at constant pressure, it produces 804 kJ of heat and does 3 kJ of work.

Answers

The enthalpy ΔH of combustion is - 804 kJ. The change in internal energy ΔE of the reaction process is -801 kJ

The objective of this question is to determine the value of ΔH and ΔE of combustion is for 1 mole of methane.

If we look closely at the question, we will realize that at constant pressure, methane burns and produces heat of 804 kJ into the surroundings.

Since heat is released;

ΔH = -804 kJ

It also does a work of 3kJ, since work is done on the system

W = - 3kJ

According to the first law of thermodynamics;

ΔE = ΔH  -  W

ΔE = (-804 - (-3 )) kJ

ΔE = -801 kJ

Therefore, we can conclude that the in 1 mole of methane, the enthalpy ΔH of combustion is - 804 kJ. The change in internal energy ΔE of the reaction process is -801 kJ

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4. The half-life of C-14 is 30 years. If a scientist had 100.0 g at the beginning, how
many grams would be left after 120 years has elapsed?
Half Lives
0
1
2
Time (years)
Amount (grams)
Show table

Answers

Answer:

6,25 gm

Explanation:

4. The half-life of C-14 is 30 years. If a scientist had 100.0 g at the beginning, how

many grams would be left after 120 years has elapsed?

10 years would be 120/30 = 4 half lives0

s0 the fraction of c-14 left would be four (1/2) multiplied together

1/2 X 1/2 X 1/2 X 1/2 = 1/16

SO 1/16 OF THE 100.0 gm would be left

100.0/16  = 6,25 gm

8) What mass of chlorine reacts with 20.0 g of iron to form iron(III) chloride?​

Answers

Answer:

38.3 g

Explanation:

8) What mass of chlorine reacts with 20.0 g of iron to form iron(III) chloride?​

first write, then balance the equation

Fe + Cl2-------------> FeCl3

2Fe + 3Cl2----------> 2FeCl3

2 moles of Fe require 3 moles of Cl2

the atomic mass of Fe is 56

20/56= 0.36 moles of Fe

this requires o.36 X 3/2 = 0.54 moles of Cl2

Cl2 has a molar mass of 35.5 X 2=71

1 mole of Cl2 is 71g

0.54 moles is 0.54 X 71 g= 38.3 g

The mass of chlorine reacts with 20.0 g of iron to form iron(III) chloride is 38.3 g.

What is mass?

Mass is defined as a numerical representation of the fundamental characteristic of all matter, inertia.

It can also be defined as a dimensionless number used to describe the mass of a particle or item.

[tex]\rm 2Fe + 3Cl_2\rightarrow2FeCl_3[/tex]

As per the reaction 3 mole of chlorine is required for 2 mole of fluorine.

Atomic mass of fluorine = 56

Moles of Fe = 20 / 56

                    =  0.36 moles

This require 0.36 x 3/2 = 0.54 moles of Cl₂

Molar mass of Cl = 35.5

So for Cl₂ = 2 x 35.5 = 71 moles

For 1 mole of Cl₂ = 71 moles is needed

So, for 0.54 moles = 71 x 0.54

                               = 38.3 g

Thus, the mass of chlorine reacts with 20.0 g of iron to form iron(III) chloride is 38.3 g.

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please help me! giving brainliest!!

Answers

Answer:

It is a covalent bond because both carbon and oxygen are nonmetals. The formula will be CO² because the electrons between the atoms are being shared equally.

Answer:

Option C

Explanation:

CO₂ is a covalent bond because we know that both elements are nonmetals. The carbon atoms shares 2 electrons with each oxygen atom and each oxygen atom shares two electrons with the carbon atom. CO₂ forms a nonpolar covalent bond, meaning that the electrons are shared equally.

Calculate the freezing point of a solution of 3.46 g of a compound, X, in 160 g of benzene. When a separate sample of X was vaporised, its density was found to be 3.27 g/Lat 116°C and 773 torr. The freezing point of pure benzene is 5.45°C, and Kis 5.12°C kg/mol.

Answers

Answer:

Calculate the freezing point of a solution of 3.46 g of a compound,X,in 160 g of benzene. When separate sample of X was vaporised, its density was found 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzen is 5.45 c, and kf is 5 .12 c/m.

Explanation:

First determine molar mass from

P*molar mass = density*R*T

Solve for molar mass. Then

moles = grams/molar mass

Solve for moles

molality = moles/kg solvent

Solve for molality

delta T = Kf*m

Solve for delta T, then convert to freezing point

The heat of combustion of propane, C3H8 (g) is -2057 kJ/mol. What would be the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas

Answers

Considering the reaction stoichiometry, the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

The balanced reaction is:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

The heat of combustion of propane, C₃H₈, is -2057 kJ/mol. This is, 2057 kJ is released for every 1 mol C₃H.

So to determine the enthalpy change if enough propane was burned to emit 12 moles of carbon dioxide, you must take into account the stoichiometry of the reaction.

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole O₂: 5 moles CO₂: 3 moles H₂O: 4 moles

Then you can apply the following rule of three: if by stoichiometry 3 moles of CO₂ are produced by 1 mole of C₃H₈, 12 moles of CO₂ are produced by how many moles of C₃H₈?

[tex]amount of moles of C_{3} H_{8} =\frac{12 moles of CO_{2}x1 mole of C_{3} H_{8} }{3 moles of CO_{2}}[/tex]

amount of moles of C₃H₈= 4 moles

So to determine the enthalpy change, you can apply the following rule of three: If for each mole of C₃H₈ 2057 kJ are released, for 4 moles of C₃H₈ how much heat is released?

[tex]Heat released=\frac{4 molesx2057 kJ}{1 mole}[/tex]

Heat released= 8228 kJ

The enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

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what are the strongest acids and bases in chemstry​

Answers

Strong Acids Strong Bases
hydrobromic acid (HBr) potassium hydroxide (KOH)
hydroiodic acid (Hl) calcium hydroxide (Ca(OH)2)
nitric acid (HNO3) strontium hydroxide (Sr(OH)2)
sulfuric acid (H2SO4) barium hydroxide (Ba(OH)2)

Mark Brainliest please

help me. correct answer will be marked as brnlst

Answers

Answer:

PCI3

Explanation:

It does not obey the octet rule on the nitrogen atom.

Place the events of a fight or flight situation in the correct order.

Question 3 options:

You spring into action

Your heart and lungs prepare to fight or run

You see the bear

Your brain sends signals to you adrenal and pituitary glands

Answers

Answer:

you see le bear

you brain sends signals

your heart and lungs receive said signals

spring to action

give some practical examples where prefixes are used

Answers

an example of a prefix is "re". this means "again". some examples of this include "retry" (to attempt again), "rehearse" (to practice for something you will later do again), and "reconsider" (to consider again).

Using the metric "stairs" convert the following:

Answers

Skuwel converter
Trust converter
Gear converter
Trigger converter

Hope it helps

What equipment is generally used to make lyophilized medications suitable for administering to the patient? a) Test tubes or sterile ampules b) Petri dishes and sterile droppers c) Sterile syringes or graduated cylinders d) Measuring cups and clean, warm water

Answers

Answer:

B) Sterile droppers and petri dishes

I will give brainlist

How much Ba(OH)2, in grams, is needed to make a 1.00m aqueous solution?

Answers

Answer:

We can say that we need 171 g barium hydroxide needed to make one model equals solution.

Explanation:

I hope this answered your question.

171 gram of  Ba(OH)2, in grams, is needed to make a 1.00 m aqueous solution.

What is mass?

Mass is the quantity of matter. The mass is calculated by the density divide by the volume.

In 1.00 solution, the mass of Ba(OH)2 is 171 gram.

Thus, the mass of Ba(OH)2 is 171 gram.

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PLEASE HELP
1. A closed bottle contains a mixture of Ar and O2 at 25.3°C. If the partial pressure of Ar in
the bottle is 0.447 atm and the total pressure of the bottle is 1.024 atm, what is the partial
pressure of O2 in the bottle?

Answers

Answer: 0.79 atm

Explanation:

The Ideal Gas Law is PV=nRT.

since both Ar and O2 have the same volume and temperature, the only variable causing the differnce in pressure betwwen the 2 gases is the number of moles

the total pressure = the partial pressure of Ar + the partial  pressure of O2

1.024 atm = 0.447 amt + ?

so  

? = 1.24 -0,447 = 1.24 - 0.45= 0.79 atm

the total pressure of the bottle is 1.024 atm, what is the partial

pressure of O2 in the bottle? is PV

Why do we need to build models of molecules to study atomic composition?

Answers

Answer:

for better understanding and visual understanding

A sample of carbon dioxide gas has a pressure of 8.33 bar and a volume of 10.3 mL at a temperature of 29.1 °C. How many
molecules of carbon dioxide gas are in the sample?
molecules:

Answers

Answer : what the rules of 8.33 because the 8.33 that is the right place for reason

Calculate energy output (ATP production) of full oxidation of saturated fatty acid with 24 carbons in the chain?
..
Explain

Answers

Correct answer………………in this picture

1) Which example shows a change in a state
of matter?
A. Liquid pouring
2
B. Wood Burning
1
C. Ice Melting

Answers

Answer:

C

Explanation:

because an ice is a solid and when it melts it becomes a liquid

Answer:

C

Explanation:

1) Which example shows a change in a state

of matter?

A. Liquid pouring

2

B. Wood Burning

1

C. Ice Melting

C  when ice melts it goes from a solid to a liquid state

Consider the following chemical equation:

C2H8 + O2 --> CO2 + H2O

Which of the following would be the best practice(s) for balancing the equation? There may or may not be more than one correct answer.

A.Reduce the coefficients if possible.

B.Put a 2 in front of the hydrocarbon.

C.Split the hydrocarbon into individual elements.

D.Split the water into HOH.

E.Keep the hydrocarbon together.

Answers

The best way to balancing a combustion equation of an organic compound, like the one in the question, is to start balancing the carbon, then the hydrogen and finish with the oxygen. For that you have to split the hydrocarbon into individual elements. And once you have finished you shall always try to reduce the coefficient if possible. Therefore, the answers are A. and C.

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