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Fofino [41]
3 years ago
8

Can anyone give me examples of control groups ?

Chemistry
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

In an experiment in which blood pressure medication is tested, one group is given the blood pressure medication while the control group is given a placebo pill.

In a test of anxiety treatment, one group attends individual therapy sessions and receives a new medication. The control group receives only an inert pill.

A treatment for drug addiction is being tested. A group of meth addicts are given the treatment while the control group is given the placebo.

Researchers are testing the effectiveness of a drug intended to reduce symptoms of Crohn's disease. Many sufferers of Crohn's disease are recruited for the effort and the group that receives the placebo is the control group.

A treatment for hair loss in men is being tested. Men of the same age range are gathered. One group receives an application with the active ingredient while the control group receives an application that appears the same but does not have the active ingredient in it.

hope this helped you

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A 35.0 mL sample of 1.00 M KBr and a 60.0 mL sample of 0.600 M KBr are mixed. The solution is then heated to evaporate water unt
Katarina [22]

Answer: The molarity of KBr in the final solution is 1.42M

Explanation:

We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.

We will first calculate the number of moles of KBr in the individual sample before mixing together

In the first sample:

Volume (V) = 35.0 mL

Concentration (C) = 1.00M

Number of moles (n) = C × V

n = (35.0mL × 1.00M)

n= 35.0mmol

For the second sample

V = 60.0 mL

C = 0.600 M

n = (60.0 mL × 0.600 M)

n = 36.0mmol

Therefore, we have (35.0 + 36.0)mmol in the final solution

Number of moles of KBr in final solution (n) = 71.0mmol

Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.

Therefore,

Final volume of solution (V) = 50mL

Number of moles of KBr in final solution (n) = 71.0mmol

From

C = n / V

C= 71.0mmol/50mL

C = 1.42M

Therefore, the molarity of KBr in the final solution is 1.42M

5 0
3 years ago
List three common counting units and their values.
emmasim [6.3K]
Dozen = 12,

ii. 1 score = 20

iii. 1 ream = 500

iv. 1 gross = 1.44
6 0
4 years ago
If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog
9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0
3 years ago
Consider this mechanism:
zaharov [31]
The answer is intermediate, so E.
6 0
3 years ago
Review the equation below. 2KClO3 mc012-1.jpg 2KCl + 3O2 How many moles of oxygen are produced when 2 mol of potassium chlorate
Vlada [557]
Stoichiometry <span>of the reaction:

</span><span>2 KClO</span>₃<span>    =    2 KCl  +   3 O</span>₂
  ↓                                      ↓
    
2 mole KClO₃ ----------> 3 mole O₂
2 mole KClO₃ ----------> ?

KClO₃ = 2 * 3 / 2

KClO₃ = 6 / 2

= 3 moles de KClO₃

hope this helps!
4 0
3 years ago
Read 2 more answers
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