Gee. I'll have to guess at what's "commonly thought".
One thing is the scale. Nobody has an accurate picture of the scale in
his head, because we never see a true-scale drawing. THAT's because
it's almost impossible to draw one on paper.
Example:
Shrink the solar system and everything in it so that the Sun
is the size of a quarter (the 25¢ coin).
Then:
-- The Earth is in orbit around the sun, 8.6 feet from it.
That's close enough that you might think you could find the
shrunken Earth. Unfortunately, it's only 0.009 inch in diameter.
-- The shrunken Jupiter is a 'huge' gas giant almost 0.1 inch in diameter.
It's orbiting the sun, about 45 feet away from it.
-- The shrunken Uranus is another gas giant, about 0.035 inch in diameter.
It's orbiting the sun, about 165 feet away from it.
-- The nearest star outside of the solar system is 441 MILES away !
On the same shrunken scale !
And there's NOTHING between here and there !
I think that's the biggest point to make about the REAL solar system ...
its utter emptiness. With the sun reduced to something you can hold
in your hand, the planets are the size of grains of sand, with hundreds
of feet of nothingness between them.
Same for its mass: The solar system is approximately nothing but a star.
That's it. A star, with some dust and some gas around it, and here and there
in the neighborhood a microscopic pebble or a chip of mineral. But mostly
it's nothing but a star ... if you went around and gathered up all that other
rubbish in the same bag and called it a part of the same solar system, the
sun would still have more than 99% of the total mass, and the bag would
hold less than 1% of it.
Book ... It's getting late, Hillary's fading, and that's all I can think of.
I hope this much is some help.
Osmotic pressure is the pressure that would have to be applied to a pure solvent to prevent it from passing into a given solution by osmosis.
That can be mathematical computed from the expression:
Osmotic pressure=C×R×T
Where,
C= Concentration
R=Gas constant
T=Temperature
Concentration=Number of moles of solute/Volume(L)
=0.005*1000/100
=0.05
R= 0.08206 atm L/mol K
T=25+273
=298
Osmotic pressure= 0.05×0.08206×298
=1.2 atm
<u>Answer:</u> 72 grams of water will be produced.
<u>Explanation:</u>
To calculate the number of moles, we use the formula:
....(1)
Mass of propane = 44 grams
Molar mass of propane = 44 grams
Putting values in above equation, we get:
For the reaction of combustion reaction of propane, the equation follows:
By Stoichiometry of the reaction,
1 mole of propane produces 4 moles of water.
So, 1 mole of propane will produce = 
of water.
Now, to calculate the amount of water, we use equation 1, we get:
Molar mass of water = 18 g/mol
Mass of water produced = 72 grams
Hence, 72 grams of water will be produced.
It doesn't?
Heat transfers from hot objects to cold objects and for ice to melt it has to increase the temperature.
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
