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4 years ago

Can you overdose or become addicted to thc or cbd?

Chemistry

2 answers:

hammer [34]4 years ago

6 0

Yes you can it depends on your body though

stealth61 [152]4 years ago

4 0

Mentally yes, but physically it cannot cause you brain to develop an addiction.

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Why males on average have VO2max than females?

katrin [286]

The average sedentary male will achieve a VO2 max of approximately 35 to 40 mL/Kg/min. And the average sedentary female will score a VO2 max of between 27 an 30 mL/Kg/min.

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3 years ago

A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?

Basile [38]

Answer:

The answer is 2.20 M

Explanation:

This is because ammonia has a pH of 11.8 and if you take 14-11.8 it equals 2.2 so the answer is 2.20 M

6 0

3 years ago

Read 2 more answers

1. Fill one container with very hot water and the other with ice water. 2. Place one end of spoon into the cup with hot water an

____ [38]

The spoon will get warmer as it takes in the heat of the water

8 0

4 years ago

What is the molarity when 0.250 moles of nacl is dissolved in water to make a 1.25 l solution?

vovikov84 [41]

Hey There!

number of moles = 0.250 moles NaCl

Volume = 1.25 L

Therefore:

M = n / V

M = 0.250 / 1.25

M = 0.2 mol/L

hope this helps!

5 0

4 years ago

The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy a

crimeas [40]

Answer:

Ea = 177x10³ J/mol

ko = $1.52x10^{19}$ J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

$k = koxe^{-Ea/RT}$

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

$\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}$

$\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}$

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x $10^{-4}$

-1.67x $10^{-5}$ xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

$k1 = koxe^{-Ea/RT1}$

$2.46 = koxe^{-177x10^3/8.314x492}$

$1.61x10^{-19}ko = 2.46$

ko = $1.52x10^{19}$ J/mol

3 0

3 years ago