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ANEK [815]
3 years ago
15

Does weed still have the same effects when put in food or drinks?

Chemistry
1 answer:
arsen [322]3 years ago
7 0
Yeah, it sure does
 
Hope this helps!


You might be interested in
How many moles are in 1.51 x 10^24 molecules of water?
lyudmila [28]

Answer:One mole of HBr has 6.02 x 1

0

23

molecules of HBr.

1 mole of HBr = 6.02 x 1

0

23

molecules of HBr.-----(a)

X mole of HBr has 1.21 x

10

24

molecules of HBr.

X mole of HBr = 1.21 x

10

24

molecules of HBr------(b)

Taking ratio of (a) and (b)

X / 1 = 1.21 x

10

24

/ 6.02 x 1

0

23

X= 2.009 moles.

Explanation:

3 0
3 years ago
What is an indication that a lake is healthy?
N76 [4]

Answer:

D

Explanation:

As bioindicators are the organism that indicate or monitor the health of the environment

7 0
3 years ago
Read 2 more answers
What is the pH of a 4.08*10-'M solution of H30+?
Mariana [72]

Answer:

8.4

Explanation:

-log(4.08x10^-9) = 8.4

- Hope that helped! Please let me know if you need further explanation.

4 0
2 years ago
Which of the following alkyl halides will react fastest with CH3OH in an SN1 mechanism?
yaroslaw [1]

Answer:

IV

Explanation:

The complete question is shown in the image attached.

Let us call to mind the fact  that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.

Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.

Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic  substitution reaction by SN1 mechanism.

5 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
2 years ago
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