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schepotkina [342]
3 years ago
9

A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i

t rises to a location where the pressure is 0.720 atm and the temperature is -12.0°C. What is the volume of the balloon
Chemistry
1 answer:
DaniilM [7]3 years ago
3 0

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

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Answer:

It is pure, anhydrous acetic acid is the correct answer.

Explanation:

  • Glacial acetic acid contains very less water that the reason it is called anhydrous.
  • Glacial acetic acid is a colorless, clear liquid and corrosive.
  • The glacial acetic acid in the diluted form used in vegetables and food preservation,flavoring in the slice, sausage, canned fruits and also it used to treat the bacteria and fungal infections.

6 0
3 years ago
A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the
cupoosta [38]

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules =T_1

Molecules helium gas = x

Moles of helium has = n_1= \frac{x}{N_A}

PV = nRT (Ideal gas equation)

PV=n_1RT_1...[1]  

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = \frac{x}{4}

Moles of helium has left after removal = n_2= \frac{x}{4\times N_A}

PV=n_2RT_2...[2]

n_1RT_1=n_2RT_2

\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2

T_1=\frac{T_2}{4}

T_2=4T_1

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

6 0
3 years ago
A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f
ratelena [41]
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
3 0
3 years ago
What is the difference between Mixtures and Compounds? I am trying to study for my Science quiz on Tuesday, and I need to know t
Deffense [45]
A mixture can be separated.  Everything in a mixture keeps it's own properties and are not chemically joined together.  I am not completely sure about the compound.  Although with the cake example, the ingredients have been mixed and kind of "fused" together upon baking. Hope this helps a little. (P.S. trail mix is a good example of a mixture.)
8 0
3 years ago
The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?
Vedmedyk [2.9K]

Answer : The volume of pure diamond is 0.493inch^3

Explanation : Given,

Density of pure carbon in diamond = 3.52g/cm^3

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}

Molar mass of carbon = 12 g/mol

\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

Density=\frac{Mass}{Volume}

Now putting all the given values in this formula, we get:

3.52g/cm^3=\frac{284.4g}{Volume}

Volume = 80.8cm^3

As we know that:

1cm^3=0.061inch^3

So,

Volume = 0.061\times 80.8inch^3

Volume = 0.493inch^3

Therefore, the volume of pure diamond is 0.493inch^3

5 0
3 years ago
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