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mihalych1998 [28]
3 years ago
14

A 8.46 L sample oxygen gas at 267K and 1.23 atm is heated to 295K. If the volume changes to 6.98L, what is the new pressure?

Chemistry
1 answer:
horrorfan [7]3 years ago
7 0

Answer:

The new pressure is 1,65 atm

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T.

(P1xV1)/T1= (P2xV2)/T2

(1,23atmx 8,46L)/267 K = (P2 x 6,98L)/ 295K

0,039 atmx L/K = (P2 x 6,98L)/ 295K

P2=(0,039 atmx L/K)x 295K/6,98L =1,65 atm

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Kaylis [27]

As one moves across a period, from left to right, both the number of protons and electrons of a neutral atom increase. The enhancing number of electrons and protons results in a greater attraction between the electrons and the nucleus. This uplifted attraction pulls the electrons nearer to the nucleus, therefore, reducing the size of the atom.  

On the other hand, while moving down a group, there is an increase in the number of energy levels. The enhanced number of energy levels increases the size of the atom in spite of the elevation in the number of protons. In the outermost energy levels, the protons get attracted towards the nucleus, however, the attraction is less due to an increase in the distance from the nucleus.  

4 0
3 years ago
What is the half life of an isotope that is 75% decayed after 16 days?
inna [77]

Answer:

half life will be 8days

5 0
2 years ago
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
2 years ago
Is thiamine mononitrate an ionic bond or covalent bond?
timofeeve [1]
It is a covalent bond. Whenever a compound uses such suffixes like mono, di, tri, tetra, and so on, it is a covalent compound- thus having covalent bonds.
7 0
3 years ago
Read 2 more answers
If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?
Ket [755]
Number of moles of FeCl2 used = mass/ molar mass 
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium 
Then 4 moles of Fe produces 8 moles of sodium.
Number of moles of sodium = mass/molar mass
Molar mass of sodium chloride = 23 +35.5 = 58.5 g/mol
Hence mass = 8 * 58.5 = 468 g. Hence Option A.

6 0
3 years ago
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