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Karo-lina-s [1.5K]
3 years ago
15

Suppose you are investigating how the amount of exercise a hamster gets affects how long the hamster lives. In order to control

the experiment, which of these should you do?
A. Give all the hamsters the exact same diet.
B. Compare how long hamsters live to how long mice live.
C. Have your friend do the same experiment with hamsters that you are doing.
D. Make all the hamsters run on a hamster wheel for the same amount of time.
Chemistry
2 answers:
ozzi3 years ago
4 0
Hello, to answer your question fully.

<span>D. Make all the hamsters run on a hamster wheel for the same amount of time.

 Because your doing the experiment on exercise not anything else 
 so keeping charts in a notebook and set a time on how long they were on it and how long you want them to be on it. 
   
                Signed by, Virtuoso Sargedog</span>

jekas [21]3 years ago
3 0

Answer:

D. Make all the hamsters run on a hamster wheel for the same amount of time.

Explanation:

Hello,

In this case, for you to study how the exercise affects the hamsters lifetime, you must put all the hamsters under the same exercising conditions, thus, by making all the hamsters run on a hamster wheel for the same amount of time, you can attain it.

Best regards.

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Atmospheric pressure arises due to the force exerted by the air above the Earth. At higher altitudes, the mass of the air above
SVETLANKA909090 [29]

Answer:

less, decreases

Explanation:

When the pressure of an atmosphere occurs because of the force exerted so at the time of the higher altitudes, the air mass i.e. above the earth should be less as the air is attracted towards surface of an earth because of the gravity and air contains the mass that shows near the surface area so automatically the air density reduced due to which the mass also decreased

7 0
3 years ago
What do you use to measure mass
Nesterboy [21]
To measure gravitational mass one of the methods includes using a spring balance. a balance is a type of scale that is different because it uses known mass to measure the unknown mass. so technically the method measures the weight.

also if you just want to know normal mass then just weigh the object on the scales because technically speaking when you want to see your weight you are actually seeing your mass. because mass never changes whilst weight is from the result of gravity and always changes, so yeah. I hope thos actually makes some sort of sense
7 0
3 years ago
A gas with a pressure of 2.25 atm occupies 450.0 mL at a temperature of 300 K. What is the volume at 405.0 K?
Aneli [31]

The volume of the gas at a temperature of 405.0 K would be 607.5 mL. Making option D the right answer to the question.

What is the volume of the gas?

To find the volume of the gas, the equation to be used would have to be combine gas law.

Combine gas law as the name suggest uses the combination of Charles law which measures Volume against temperature, and Gay-Lussac's law which measures Pressure/Temperature, and Boyle's law  which measures pressure X volume where k is constant.

Using the combine law to find the volume, we have:

P₁V₁/T₁=P₂V₂/T₂

Where P₁ = initial pressure

           V₁ = initial volume

             T₁ = initial temperature

            P₂ = final  pressure

            V₂ = final  volume

            T₂ = final  temperature

P₁ = 2.25atm

V₁ = 450.0 mL

T₁ = 300 K

T₂ = 405.0 K

V₂ = ?

D) 607.5 mL

= [2.25(450)]÷300=[2.25(V₂]÷405

Making  V₂ the subject

 3.375=2.25 V₂ ÷ 405

V₂ = 3.375 x 405 ÷ 2.25

V₂ = 607.5 mL

In summary, a gas with an initial pressure of 2.25atm, an initial pressure of 450.0 mL and an initial temperature of 300 K would have a final volume of 607.5 mL if the temperature is increased to  405.0 K.

Learn more about Combine gas law here: brainly.com/question/13538773

#SPJ1

6 0
1 year ago
How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
10.1 g O2 left over</span><span>
</span>
7 0
3 years ago
Read 2 more answers
When a 3.00 g 3.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 1.36 kJ ⋅ K − 1
cupoosta [38]

Answer:

Molar heat of solution of KBr is 20.0kJ/mol

Explanation:

Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.

The dissolution of KBr is:

KBr → K⁺ + Br⁻

In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:

q = 1.36kJK⁻¹ × 0.370K

q = 0.5032kJ

Moles of KBr in 3.00g are:

3.00g × (1mol / 119g) = 0.0252moles

Thus, molar heat of solution of KBr is:

0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>

3 0
3 years ago
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