I think it's aluminium chloride
Answer:
a magnifying scope
Explanation: you get a close up view of the target and because of that there is a lower margin of error
Answer: <u><em>Shape and position</em></u>
Explanation:
<u><em>I just took the test and I got it correct</em></u>
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.
Answer:
2.14 moles of H₂O₂ are required
Explanation:
Given data:
Number of moles of H₂O₂ required = ?
Number of moles of N₂H₄ available = 1.07 mol
Solution:
Chemical equation:
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
now we will compare the moles of H₂O₂ and N₂H₄
N₂H₄ : H₂O₂
1 : 2
1.07 : 2×1.07 = 2.14 mol
