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3 years ago

Define Condensation .

Chemistry

2 answers:

ozzi3 years ago

6 0

Answer:

Condensation is the change of the physical state of matter from the gas phase into the liquid phase.

Rom4ik [11]3 years ago

3 0

Answer:

the conversion of a vapour or gas to a liquid is defined as condensation.

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a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa

lesya692 [45]

Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

Where:

V₁ = Initial volume

V₂ = Final volume = V₁/2

P₁ = Initial pressure = 832 torr

P₂ = Final pressure = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

6 0

3 years ago

Read 2 more answers

Why do substances burn with or without flame?

Gala2k [10]

Flame (fire) is the effect of a chemical reaction that produces visible light and heat. The chemical reaction is going on in the substance being burned.. Thats why coals glow and flames seem to leap into the air.

If your reaction does not have a flame, then either it is not producing visible light or the reaction does not occur in the air above the substance.

7 0

3 years ago

Read 2 more answers

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

C4H10 + O2----------------> CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A

8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm

6 0

3 years ago

Explain how plantesimals differ from protoplanets

MAXImum [283]

Answer:

Planetesimals are solid objects thought to exist in protoplanetary disks and in

A protoplanet is a large planetary embryo that originated within a protoplanetary disc and has undergone internal melting to produce a differentiated interior.

5 0

3 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 "burn", 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 conversión factor 1 L = 1000 cm³.

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

Since density = mass/volume, 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 114 g octane : 12.5 moles of oxygen

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

3 years ago

Define Condensation .​

Define Condensation .