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Sindrei [870]
3 years ago
10

PLEASE HELP! Class is almost ending, I don't want to repeat.

Chemistry
1 answer:
nikklg [1K]3 years ago
6 0
ANSWERS:

1. 2KClO_{3} ⇒ 2KCl + 3O_{2}

2. H_{2}  C_{2} O_{4}  + 2NaOH ⇒ Na_{2}  C_{2} O_{4}  + 2 H_{2} O

3. 2 C_{10}  H_{22} +  31O_{2} ⇒ 20C O_{2} + 22 H_{2} O

4. 4 H_{2} + Fe_{3} O_{4} ⇒ 3Fe +4 H_{2} O

5. 4P + 5 O_{2} ⇒ 2 P_{2}  O_{5}


I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

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PLEASE HELP ME BUT FAST!
Elina [12.6K]
Hey, i add an answer so you can give the other person brainlest :)
7 0
3 years ago
Read 2 more answers
What is the molarity of 3.2 moles of magnesium chloride in 1.5 L of water
diamong [38]

Answer:4.8

Explanation:

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3 years ago
Which type of electromagnetic wave has the greatest wavelength?(1 point)
Kruka [31]
The answer is A ……..
5 0
2 years ago
Need help !!!!! ASAP
Julli [10]
<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

V_{2}=1L

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

P_{1}V_{1}=P_{2}V_{2}

So, we are given the information:

V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa

Then, isolating the new volume and substituting into the equation, we have:

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\frac{P_{1}V_{1}}{P_{2}}

V_{2}=\frac{50kPa*2L}{100kPa}=1L

Hence, the new volume will be 1 L.

V_{2}=1L

Have a nice day!

4 0
2 years ago
BEST GETS BRAINLIEST!!
tatuchka [14]
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.
3 0
3 years ago
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