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kolbaska11 [484]
3 years ago
14

Controls are defined as____

Chemistry
1 answer:
Alexeev081 [22]3 years ago
5 0
A scientific control is an experiment or observation designed to minimize the effects of variables other than the independent variable.
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Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
3 years ago
What is the difference between the plasma and gaseous states of matter?
Ivenika [448]
In gaseous particles are slightly lighter than those in pasma which look like liquid
8 0
3 years ago
Compared with the fibers of cotton plants growing today, what is the relative ratio radioactivity in the old material vs the rel
TiliK225 [7]

Answer:

0.56

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/At)

t1/2 = half life of the C-14 = 5730 y

t = time elapsed = 4800 y

At = Activity of C-14 at time t

Ao= Activity of a living C-14 sample

0.693/5730 = 2.303/4800 log (Ao/At)

1.2 * 10^-4 = 4.8 * 10^-4 log (Ao/At)

log (Ao/At) = 1.2 * 10^-4/4.8 * 10^-4

log (Ao/At) = 0.25

Ao/At = Antilog (0.25)

Ao/At = 1.778

Hence;

At/Ao = (1.778)^-1

At/Ao = 0.56

7 0
3 years ago
En un recipiente colocamos unos cubos de hielo luego tapamos y observamos que sucede despues de un tiempo
SSSSS [86.1K]

Three questions come along with the given statement. It is in Spanish language:



a) Por qué se humedeció la parte exterior del frasco?


b) Por qué el hielo disminuyó su volumen y ahora es agua?


c) Por qué puede haber agua en el exterior?



These are the three answers (in English).



First question:



a) Por qué se humedeció la parte exterior del frasco?



The question is Why did the outside of the bottle get wet?



Answer:



The outside of the bottle get wet because the ice cubes cooled the walls of the bottle, so the air surrounding the bottle also cooled.



The air contains humidity (water) in gas phase. The hotter the air the more the amount of humidity it can retain, the cooler the air the less the amount of humidity it can retain.


Then, when the air close to the walls of the bootle got cooler some of the water in the air became liquid and those are the drops of water that you see in the outside of the bottle.



Second question



b) Por qué el hielo disminuyó su volumen y ahora es agua?



The question is Why did the ice diminish its volume and now it is water?



Answer:




The ice diminished its volume and now it is water, becasue the ice, which is cooler than the surroundings, received heat energy (from the surroundings) and then its temperature increased. At some moment, this temperature reached the melting point of the ice (water) and it started to become liquid.



Third question



c) Por qué puede haber agua en el exterior?



The question is: Why can there be water outside?




Answer:



The water outside is outside since the beginning: it is in the air. You do not see it because it is gas state. When the air close to the walss of the bottle got cooler, part of the water in the air became liquid.

4 0
3 years ago
Define the term macromolecule
makkiz [27]

Explanation:

a molecule containing a very large number of atoms, such as a protein, nucleic acid, or synthetic polymer.

4 0
2 years ago
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