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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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Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react
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Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of BCl_3 = 60 g

Mass of H_2O = 37.5 g

Molar mass of BCl_3 = 117 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of HCl = 36.5 g/mole

First we have to calculate the moles of BCl_3 and H_2O.

\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)

From the balanced reaction we conclude that

As, 1 mole of BCl_3 react with 3 mole of H_2O

So, 0.513 moles of BCl_3 react with 3\times 0.513=1.539 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and BCl_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of HCl.

As, 1 mole of BCl_3 react to give 3 moles of HCl

So, 0.513 moles of BCl_3 react to give 3\times 0.513=1.539 moles of HCl

Now we have to calculate the mass of HCl.

\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl

\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g

Therefore, the theoretical yield of HCl is, 56.1735 grams

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