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Nat2105 [25]
3 years ago
14

a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa

ctor of 2 what will happen to the pressure? you may assume the temperature remains at 35 c
Chemistry
2 answers:
lesya692 [45]3 years ago
6 0

Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

Where:

V₁ = Initial volume

V₂ = Final volume = V₁/2

P₁ = Initial pressure = 832 torr

P₂ = Final pressure  = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂  = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

Nonamiya [84]3 years ago
4 0

Answer:

The pressure will increase by a factor of 2 and is now 1664 torr

Explanation:

The question says for us to assume that temperature is constant. Now, since we are given pressure and volumw, we will use Boyle's law which is a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature i.e V ∝ 1/P

Thus, PV = K

Where K is a constant.

So,

P₁•V₁ = P₂•V₂ = k

Where:

P₁ = Initial pressure

V₁ = Initial volume

P₂ = Final pressure

V₂ = Final volume = V₁/2

From the question, P₁ = 832 torr ; we are told that volume is decreased by 2,thus, V₂ = V₁/2

Now, we want to find out what will happen to the pressure P₂.

Let's make P₂ the subject;

P₁•V₁ = P₂•V₂

Thus, P₂ = (P₁•V₁)/V₂

Plugging in the relevant values to obtain ;

P₂  = (P₁•2V₁)/V₁

V₁ will cancel out and we have;

P₂  = 2P₁

Now, we are given that P₁ = 832 torr, Thus,

P₂ = 2P₁ = 2× 832 torr = 1664 torr

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zheka24 [161]

Answer:

Charge the balloon, hold it near an electroscope, and determine if the electroscope leaves move.

Explanation:

The gold leaf electroscope is an instrument used to detect if a body is charged. It has two gold leafs suspended from a brass stem in a vacuumed glass jar and connected to a metal cap(Toppr).

When the test body is allowed to touch the metal cap, a change in the size of the leaves shows whether the body is charged or not.

Since we are suspecting the balloon to be made up of a metal; metals can be charged. We can test if there is really a charge on the balloon by bringing it near an electroscope to see if the electroscope moves.

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3 years ago
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4 0
3 years ago
If the temperature we're to go down how would that effect the solubility of each solute
vova2212 [387]

well when there is more kinetic energy in a solution the higher its temperature is going to be so the more the molecules are able to be help together, the lower the temperature the more difficult it would be to break the molecule apart in order to bond with it.

6 0
3 years ago
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
A piece of metal with a mass of 114 g was placed into a graduated cylinder that contained 25. 00 ml of water, raising the water
alexgriva [62]

The density of the metal is 6.51 g/cm3 .

<h3>What is density, for instance?</h3>

How much "stuff" is contained in a specific quantity of space is determined by its density. For instance, a block of the harder, lighter material gold (Au) will be denser than a block of the heaviest element lead (Pb) (Au). Styrofoam blocks are less dense than bricks. Mass per unit size serves as its definition.

Briefing :

mass, m = 114 g

initial volume, V1 = 25 mL

final volume, V2 = 42.5 mL

Volume of the metal piece, V = V2 - V1

                                                 = 42.5 - 25

                                                 = 17.5 mL

1 mL = 1 c.c

So, Volume of metal, V = 17.5 c.c.

Let the density of the metal is d.

density = mass / volume

         d = 114 / 17.5

            = 6.51 g/c.c

Thus, the density of metal is 6.51 g/c.c.

To know more about density visit ;

brainly.com/question/15164682

#SPJ4

3 0
1 year ago
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