Ask question

Ask question

All categories

3 years ago

6

How is the Methane on Titan similar to water on Earth?

2 answers:

spin [16.1K]3 years ago

6 0

Both in the phase of a liquid & has similar qualities like both molecules.

IceJOKER [234]3 years ago

5 0

Saturn's largest moon Titan is the only world in our solar system besides Earth known to have bodies of liquid on its surface. ... It's now known that Titan's hydrologic cycle is surprisingly similar to Earth's, with one big exception: the liquid on Titan is liquid methane/ethane instead of water, due to the extreme cold.

You might be interested in

What is a contributor of global warming?

iogann1982 [59]

Carbon dioxide emissions from, well, whatever emits CO2 really.

8 0

3 years ago

Read 2 more answers

Q1. If a piece of ice absorbs 131 J of heat energy, how much was lost by the surroundings?

Rashid [163]

Answer:

53

Explanation:

4 0

3 years ago

If 15 L of chlorine reacts at a constant temperature of 298 K and pressure of 9 atm, how many grams of hydrochloric acid (HCl) a

stepan [7]

Answer:400k

Explanation:

7 0

3 years ago

Plz help me well mark brainliest if you are correct!

lara [203]

Answer:

c: heat and erosine

Explanation:

5 0

3 years ago

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

Mademuasel [1]

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 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 propane at $25.0^oC$ = ?

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

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

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

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

3 0

3 years ago