Chemistry, 25.06.2019 12:00 902coo

18.a 2.50 l sample of dry air in a cylinder exerts a pressure of 3.00 atm at a temperature of 25 c. the cylinder is reduced to 1.00 atm. what is the volume of the gas without a change in temperature?

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Answer from: volleyballfun24

Taking into account the Boyle's law, the volume of the gas without a change in temperature will be 7.50 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.

Boyle's law is one of the gas laws, which states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: that is, if the pressure increases, the volume decreases ; whereas if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as the product of pressure times volume equal to a constant:

P×V= k

Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:

P1×V1 = P2×V2

In this case, you know:

P1= 3 atmV1= 2.50 LP2= 1 atmV2= ?

Replacing in Boyle's law:

3 atm× 2.50 L= 1 atm× V2

Solving:

$V2=\frac{3 atmx2.50 L}{1 atm}$

V2= 7.50 L

Finally, the volume of the gas without a change in temperature will be 7.50 L.

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Answer from: tyreshagarrett123

$\boxed{\text{7.50 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{3.00 atm}\qquad & V_{1} &= & \text{2.50 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}3.00 \times 2.50 & =& 1.00V_{2}\\7.500 & = & 1.00V_{2}\\V_{2} & = &\textbf{7.50 L}\\\end{array}\\\text{The volume of the gas is } \boxed{\textbf{7.50 L}}$

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.