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Anastaziya [24]
3 years ago
6

A lab is done using different metals and the density of the metals are recorded in a data table. What is the independent variabl

e in the lab?
Chemistry
1 answer:
vovikov84 [41]3 years ago
6 0

Answer:   The independent variable is the type of metal being used.

{Note:  The "dependent variable" is the "measured density" that corresponds to each of the metals."}.

___________________________________________

Explanation:

___________________________________________

The "independent variable", which is plotted on the "x-axis" (horizontal axis), is the variable that can be "controlled/manipulated".  In this case, this would be the type of metal chosen.

The "dependent variable" , which is plotted on the "y-axis" (vertical axis) is the "obtained value/measurement/result" (that "cannot be controlled/manipulated").

In this case, the "density", which is the "measured value" that corresponds to the selected "meal", is the "dependent variable".

___________________________________________

Hope this helpful to you!

   Wishing you well!

___________________________________________

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4. If an experimental value for a given observation was 415 nm while the theoretical value
Alex

Answer: % error of observation is 4.77%

Explanation:

Given:

Observation value = 415nm

theoretical value= 435.8nm

Percent error of observation = theoretical value- observation value/ theoretical value x 100 %

= 435.8-415/435.8= 0.04772 x 100 = 4.77%

therefore % error of observation is 4.77%

7 0
3 years ago
Draw the Lewis structure for carbonate
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8 0
3 years ago
Base A dissociates 25% in water. Base B dissociates 50%. Base C dissociates
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Answer:Base A is the weakest conductor electricity

Explanation:

Dissociation is a factor that affects electrical conductivity. The greater the percentage of dissociation for bases the stronger the conductivity of electricity.

Given that

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We can conclude that  Base A is the weakest conductor oelectricity since it has the lowest percentage of dissociation.

5 0
2 years ago
Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm
pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

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145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
In the chemical reaction, calcium carbonate (CaCO3) was heated to form two new
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Answer: 10.99

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