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rosijanka [135]
3 years ago
13

0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?

Chemistry
1 answer:
ioda3 years ago
4 0

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

You might be interested in
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
3 years ago
Where would a primary producer be located in a food chain?
vfiekz [6]

Answer:

The primary producer would be at the bottom of the food chain.

Explanation:

4 0
3 years ago
Read 2 more answers
Calculate the amount of heat required to raise the temperature of a 24 g sample of water from 9°C to 23°C.
borishaifa [10]

Answer:

1400KJ/mol⁻¹

Explanation:

Amount of heat required can be found by:

Q = m × c × ΔT

<em>Where m is the mass, c is the specific heat capacity (4.2KJ for water) and ΔT is the change in temperature.</em>

Q = 24 × 4.2 × (23 - 9)

= 24 × 4.2 × 14

=   1411.2KJ/mol⁻¹

= <u>1400KJ/mol⁻¹</u>  (to 2 significant figures)

7 0
3 years ago
At S?fndnfnfnfxmckfkffkfkfk ckfkfkfkfkfkv kfflfmcmc
IRINA_888 [86]

Answer:

what?

Explanation:

3 0
3 years ago
What is Y? 222/86 Rn A/z + 4/2He
kondaur [170]

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}

Y is Po-218.

  • A = 218
  • Z = 84.
<h3>Explanation </h3>

^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}

Here's the symbol of a particle in a nuclear reaction. ^{A}_{Z}\text{Y}.

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

  • The mass number on the left hand side is 222.
  • The two mass numbers on the right hand side add up to A + 4.
  • 222 = A + 4.
  • A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

  • The atomic number on the left hand side is 86.
  • The two atomic number on the right hand side add up to __ + __.
  • 86 = __ + __.
  • Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as ^{218}_{\phantom{2}84}\text{Po}. Hence the equation:

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}.

3 0
3 years ago
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