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PIT_PIT [208]
3 years ago
13

The gas in a 600. mL balloon has a pressure of 1.20 atm. If the temperature remains constant, what will be the pressure of the g

as in the balloon when it is compressed to 400. mL?
Chemistry
1 answer:
enyata [817]3 years ago
8 0

Answer:

1.8 atm

Explanation:

As temperature and moles of a gas reamin both constant, we can say that:

Pressure₁ . Volume₁ = Pressure₂ . Volume₂

1.20 atm . 600 mL = Pressure₂ . 400 mL

(1.20 atm . 600 mL) / 400 mL = Pressure ₂ → 1.8 atm

You might be interested in
What is the empirical formula for c12h24o6? what is the empirical formula for c12h24o6? ch2o cho c2h5o c2h4o cho2?
stira [4]

Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.

Now, molecular formula is equal to the product of n (ratio) and empirical formula.

Molecular formula = n\times empirical formula    (1)

molecular formula =C_{12}H_{24}O_{6} (given)

Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.

empirical formula becomes C_{2}H_{4}O

Thus, according to the formula (1)

C_{12}H_{24}O_{6} = 6\times C_{2}H_{4}O

Hence, empirical formula of given molecular formula is C_{2}H_{4}O


4 0
3 years ago
Drag each tile to the correct location on the image.
nikdorinn [45]

Answer:

[He]: 2s² 2p⁵.

[Ne]: 3s².

[Ar]: 4s² 3d¹⁰ 4p².

[Kr]: 5s² 4d¹⁰ 5p⁵.

[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².

Explanation:

  • Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.

  • [He]:

He contains 2 electrons fill 1s (1s²).

So, [He] can be written before the electronic configuration of 2s² 2p⁵.

  • [Ne]:

Ne contains 10 electrons fill (1s² 2s² 2p⁶).

So, [Ne] can be written before the electronic configuration of 3s².

  • [Ar]:

Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).

So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².

  • [Kr]:

Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).

So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.

  • [Xe]:

Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).

So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².

3 0
4 years ago
Read 2 more answers
A cylinder is labeled \"PENTANE.\" When the gas inside the cylinder is monochlorinated, five isomers of formula C5H11Cl result.
melisa1 [442]

Answer:

a mixture of two these

Explanation:

The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.

n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.

Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.

Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.

Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.  

8 0
3 years ago
What is the maximum amount in moles of P2O5 that can theoretically be made from 235 g of P4 and excess oxygen?
Lady_Fox [76]
The correct answer would be 3.8mol
5 0
3 years ago
Which of the following shows the conservation of mass during cellular respiration? 3 CO2 3 H2O energy → 3 C6H12O6 3 O2 6 CO2 6 H
Thepotemich [5.8K]

The reaction that has been, following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

The law of conservation has been given in the chemical reaction that there has been no loss or gain of the mass and energy.

The law of conservation has been evident when there has been an equal number of atoms of each element on the product and the reactant side.

<h3 /><h3>Conservation of mass in Cellular respiration</h3>

The following reactions have been identified as:

  • \rm 3\;CO_2\;+\;3\;H_2O\;+\;Energy\;\rightarrow\;3\;C_6H_1_2O_6

Carbon atoms

Reactant = 3

Product = 18

Oxygen atoms

Reactant = 9

Product = 18

Hydrogen atoms

Reactant = 6

Product = 36

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm 3\;O_2\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Energy\;\rightarrow\;C_6H_1_2O_6\;+\;6\;O_2

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 24

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 18

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is equal on the product and reactant side, thus follows the law of conservation of mass.

  • \rm 6\;H_2O\;+\;C_6H_1_2O_6\;\rightarrow\;6\;O_2\;+\;Energy

Carbon atoms

Reactant = 6

Product = 0

Oxygen atoms

Reactant = 12

Product = 12

Hydrogen atoms

Reactant = 24

Product = 0

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

Thus, the reaction that has been following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

Learn more about law of conservation, here:

brainly.com/question/2175724

4 0
2 years ago
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