Is it always easy to classify research as applied or pure? Explain.

What is energy being added to the water cycle?

Komok [63]

Answer:

yes

Explanation:

6 0

3 years ago

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,

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

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

3 years ago

Giving brainly if correct

mel-nik [20]

Answer:

The corect answer would be C.

Explanation:

The flow rate set at a differentt time would be the correct measurement beecause wats and speed add up to your main answer.

3 0

2 years ago

Read 2 more answers

An air bubble with a volume of 5.0 ml is released at the bottom of a lake where the pressure is 3.0 atm. when it reaches the sur

Svetradugi [14.3K]

The new volume of the air bubble that has an initial volume of 5.0 ml released at the bottom of a lake where the pressure is 3.0 atm is 15mL.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated by using the following formula:

P1V1 = P2V2

Where;

P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume

5 × 3 = 1 × V2

15 = V2

V2 = 15mL

Therefore, the new volume of the air bubble that has an initial volume of 5.0 ml released at the bottom of a lake where the pressure is 3.0 atm is 15mL.

Learn more about volume at: brainly.com/question/1578538

7 0

2 years ago

As the temperature of a material increases, the average __(blank)__ of its particles increases.

AnnZ [28]

2 it’s the correct answer

7 0

3 years ago