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svetlana [45]
3 years ago
8

How does the control group differ from the experimental group? Choose all the correct options.

Chemistry
2 answers:
Marianna [84]3 years ago
5 0

Answer:

B,D,E

Explanation:

emmainna [20.7K]3 years ago
3 0
The answer is B,D,E those are the correct options.
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Describe several main ideas in chemistry
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Matter.

A force of attraction that holds atom together 
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3 years ago
Convert 121 Cal to kilowatt-hours
gulaghasi [49]

Answer:

here

Explanation:

0.000141 to kilowatt-hours. hope this helped

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3 years ago
Why would dry cleaners use nonpolar tetrachloroethylene, C2Cl4, to get cheeseburger grease out of your clothes?
Kobotan [32]

It is water-soluble. hope this helps


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3 years ago
Read 2 more answers
What is the entropy change of the surroundings
KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

5 0
3 years ago
Sodium metal is sometimes used as a cooling agent in heat exchange units because of its relatively high molar heat capacity of 2
IgorC [24]

Answer:

The specific heat of sodium is 1,23J/g°C

Explanation:

Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.

starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C

28,2 \frac{J}{mol C} x \frac{1mol}{23g} = 1,23 J/g C

4 0
3 years ago
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