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aliina [53]
3 years ago
8

Compare the density of gases in the thermosphere to the density at sea level.

Chemistry
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

The atmosphere is 10 million times

Explanation:

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The temperature and number of moles of a gas are held constant. Which of the following is true for the pressure of the gas?
Taya2010 [7]

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K         P2V2 = K

P1V1 = P2V2

3 0
4 years ago
When a solution forms, what interactions between particles are involved? Which are exothermic? Which are endothermic?
Blizzard [7]

Answer:

Explanation:

There are three types of interactions involved between the particles when solution are formed.

1 : Solute - solute interaction:

2 : Solute - solvent interaction:

3 : Solvent - solvent interaction:

1 : Solute - solute interaction:

It is the inter-molecular attraction between the solute particles.

2 : Solute - solvent interaction:

It involve the inter-molecular attraction between solvent and solute particles.

3 : Solvent - solvent interaction:

It involve the intermolecular attraction between solvent particles.

Solutions are formed if the intermolecular attraction between solute particles are similar to the attraction between solvent particles.

Exothermic process:

The process will exothermic when solute solvent bonds are formed with the release of energy and energy required to brake the solute-solute particles and solvent solvent particles are less.

Endothermic process:

The process will be endothermic when energy required to break the solute-solute particles and solvent solvent particles are higher than energy released when solute solvent bonds are formed .

6 0
3 years ago
Will name brainliest!!!!!
dem82 [27]

According to sources, the most probable answer to this query is the reaction is spontaneous and will proceed without any energy input. This means that energy was produced and can result to a reaction.We appreciate your questions. Please, never hesitate to ask more in Brainly your queries. 
7 0
3 years ago
Read 2 more answers
please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi
Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>
5 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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