Answer:

Explanation:
Hello,
In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

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Answer:
Q = 114349.5 J
Explanation:
Hello there!
In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Thus, the total energy turns out to be:

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The two notations that represent isotopes of the same element is the one that represented in option 1
The lower number is the number of protons while the upper number is the atomic weight
hope this helps
Answer: Plastic water bottles
Explanation:
If you use disposable water bottles, here are some important concerns you should know about how they’re made as well as the problems they cause for the planet, your health, and your wallet.