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hammer [34]
3 years ago
8

Where are chemicals found?

Chemistry
1 answer:
ziro4ka [17]3 years ago
4 0
Chemicals are found in all things. Solids, liquids, and gases.
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Which of the following would contain 48 electrons?
Helga [31]
The answer is cadmium its got 48 electrons which is y its number 48 on the period table
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WHICH STATE OF MATTER IS THIS: PARTICLES VIBRATE AND CHANGE POSITION
SVETLANKA909090 [29]

Answer: Gas. Gas vibrates and move freely at high speeds.

Explanation:

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3 years ago
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
Which element probably reacts most like bromine, Br?
Nikitich [7]
D) Chlorine, Cl. Hope that helped
8 0
3 years ago
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Could volume by displacement be used to determine the volume of a lump of rock salt? Explain would be nice
Readme [11.4K]

Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:

1).  Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water.  So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease.  If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.

So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve.  I don't know right now what that could
be.  You'd have to shop around and find one.

2).  Whatever fluid you did choose, it would also have to be less dense
than rock salt.  If it's more dense, then the rock salt just floats in it, and
never goes all the way under.  If that happens, then you have a tough
time measuring the total volume of the lump.

So the displacement method could perhaps be used, in principle, but
it would not be easy.


5 0
3 years ago
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