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erastovalidia [21]
3 years ago
9

After developing your hypothesis, the next step in the inquiry process is to design a controlled experiment in order to test you

r hypothesis. In a controlled experiment, all variables are kept constant except the one that you are testing. The variables you keep the same are called controlled variables. The variable that you change or test is called the manipulated variable. The variable that responds to changes you make is called the responding variable. a. How would you design a controlled experiment to test the hypothesis you made about the volume of a gas? b. What is the manipulated variable in your experiment? c. What is the responding variable in your experiment?
Chemistry
1 answer:
Roman55 [17]3 years ago
6 0
For your experiment, you would need to see decide what your hypothesis is before you design the experiment. If you want to see if the volume of gas changes depending on the container it's in, then you need to set it up depending on what you want to do with the volume.
The manipulated variable would be whatever you change, so if you switch container sizes, that would be the manipulated variable and the responding variable would be the volume of the gas.

I hope this helps!
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Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.​
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Answer:

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

Explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl = \frac{14.5}{58.5} =  0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

=  1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

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<h3> I hope it'll help you....</h3>
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