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4 years ago

13

At which temperature and pressure will a sample of neon gas behave most like an ideal gas?

2 answers:

Andreas93 [3]4 years ago

4 0

Answer:

At STP, 760mmHg or 1 atm and OK or 273 degrees celcius

Explanation:

The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.

This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas

m_a_m_a [10]4 years ago

3 0

The given question is incomplete. The complete question is as follows.

At which temperature and pressure will a sample of neon gas behave most like an ideal gas?

Choices are as follow:

(1) 100 K and 0.25 atm

(2) 100 K and 25 atm

(3) 400 K and 0.25 atm

(4) 400 K and 25 atm

Explanation:

At low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas. Hence, gases behave ideally at these conditions.

Whereas at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, there exists force of attraction between the molecules at low temperature and high pressure and under these conditions gases are known as real gases.

For the given conditions, 400 K and 0.25 atm depicts low pressure and high temperature.

Thus, we can conclude that at 400 K and 0.25 atm a sample of neon gas behave most like an ideal gas.

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Balance the equation: Ca + ....... H2O->Ca(OH)2 + H2

Mnenie [13.5K]

$\large\boxed{{Ca+ \bold{2}H_2O \to Ca(OH)_2+H_{2}}}$

7 0

4 years ago

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc

White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says <em>"burn"</em>, it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

55 L of gasoline (assuming gasoline to be pure octane).
The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the <em>conversión factor 1 L = 1000 cm³</em>.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since <em>density = mass/volume</em>, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

For octane:

molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

12 g/mol . 8 + 1g/mol . 18 = 114 g/mol

According to the balanced equation reacts 1 mol of octane, which means 114 grams of it.

For oxygen:

According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is <u>114 g octane : 12.5 moles of oxygen</u>

<u />

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$

4 0

4 years ago

What can you conclude from the figure shown?

arlik [135]

Answer:

A. Hund's rule has been violated

Explanation:

There must be one electron with the same spin in each orbital of the same energy before you can put two in the same orbital. In the photo, the 2s sublevel is completely filled before the 1s sublevel (if anything the 1s should have an up spin and down spin, with the 2s having one up spin).

6 0

3 years ago

A solution that has a pOH of 5.36 is-

sdas [7]

Answer:

Option D

Explanation:

A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.

A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale.

The hydroxide ion molarity may be expressed as a p-function, or pOH.

pOH = −log[OH−]

Basic solutions are those with hydronium ion molarities less than 1.0 × 10−7 M and hydroxide ion molarities greater than 1.0 × 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).

7 0

3 years ago

Period 6, group 12 (2B)

maksim [4K]

Answer:

mercury or Hg

Explanation:

i looked at the periodic table haha

hope this helps and is right <3

6 0

4 years ago