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Rasek [7]
3 years ago
11

Scientists are testing a new drug, which they think might allow people to memorize large amounts of information. Group A is taki

ng a sugar pill. Group B is taking the new drug. Both groups are 25–35 years old and are not taking any other medicine. After a month, both groups will be given a memorization test, and scientists will compare each group’s ability to memorize new information. Which choice is the controlled variable (constant) in the experiment? A. the sugar pill B. the new drug C. the ability to memorize D. the age of the test subjects
Chemistry
2 answers:
musickatia [10]3 years ago
6 0
The answer to this question is C: the ability to memorize

Hope this helps :)
erastovalidia [21]3 years ago
5 0
The answer is C.the ability to memorize. Hope I am not late.
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Consider the statement, “In a chemical reaction, atoms are neither created nor destroyed, only rearranged.” Why is this a statem
Karo-lina-s [1.5K]

Answer: The law of conservation of mass is reffering to the fact that energy cannot be created or destroyed. So when you speak about atoms not being created or destroyed it is the same thing. Unless you're talking about an atomic bomb where the atoms are split.

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Helen adjusts the armature of an electric generator by increasing the number of coils around the iron core. what is helen most l
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Increase the amount of current

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3 years ago
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1. Three
Bess [88]

Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

5 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
2 years ago
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