Think about a hot summer afternoon at the beach

Chemistry

2 answers:

pentagon [3]2 years ago

4 0

Answer:

Explanation:

I say this because of this that I read. There was a study shown

"Our experiment was a success. We obtained data that answered our problem, “Which substance absorbs the most heat on a hot day? Sand, water, or sand with water?” The results showed that from the twenty-five trials: the cup of sand temperature ranged from 63-68°C, cup of water temperature ranged from 71-77°C, and the cup of sand with water temperature ranged from 66-71°C. Based on our data, our hypothesis was incorrect. We hypothesized that the sand would absorb the most heat since it is darker and less reflective than water; but from our data, water absorbed the most heat. In conclusion, sand absorbed the least amount of heat and water absorbed the most. Next time we want to cool down at the beach, we’ll know where to go!"

This was an experiment that was performed and this is the conclusion

Hope that helps

NeTakaya2 years ago

3 0

A. Would be the correct answer :-)

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A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$