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umka21 [38]
3 years ago
5

A small sample of an organic compound (a nonelectrolyte) was added to benzene (Freezing point = 5.5°C)

Chemistry
1 answer:
Alja [10]3 years ago
3 0

The new lowered freezing point of the benzene mixture is determined as 2.4°C.

<h3>Effect of impurities in freezing point</h3>

The presence of impurities lowers the vapour pressure of the solution since the concentration of the solution is increased.

Change in temperature of the new mixture = 0.613 x 5.5°C = 3.1 °C

New freezing point = 5.5°C - 3.1°C = 2.4°C

Thus, the new lowered freezing point of the benzene mixture is determined as 2.4°C.

Learn more about freezing point here: brainly.com/question/24314907

#SPJ1

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Answer:  I say its Segmented body  and radial symmetry im not sure if it is right

Explanation:

3 0
3 years ago
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How many moles of carbon are in a sample of 21.45 moles of heptane(C7H16)???????
saw5 [17]

The  moles  of carbon  that  are in the sample  of 21.45  moles of heptane(C₇H₁₆)  is 150.15   moles

  <u><em>calculation</em></u>

 moles  of carbon = moles  of  heptane  × number of C atom

 number of C atom  in heptane = 1 ×7  = 7 atoms

moles is therefore = 21.45 moles ×  7 =150.15  moles

3 0
3 years ago
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A superconductor performs best at??
Charra [1.4K]

very cold temperatures

Explanation:

A superconductor performs best at very cold temperatures.

A superconductor is a perfect conductor that is able to allow the passage of electricity and heat without resistance.

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Metals brainly.com/question/2474874

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4 0
3 years ago
How much energy, in joules, is released when 17 kilograms of granite is cooled from 45 C to 21 C?
Lostsunrise [7]
Answer is: <span>D. 327,992.8 J.
</span>m(granite) = 17 kg = 17000g.
ΔT(granite) = 21°C - 45°C = -24°C (-24K).<span>
cp(granite</span>) = 0,804 J/g·°C, <span>specific heat capacity of granite.
Q = m(granite</span>) · ΔT(granite) · cp(granite).<span>
Q = 17000 g ·(-24</span>°C)<span>· 0,804 J/g·K.
Q = -327990 J.
</span>The granite lost 327990 joules of energy.<span>
Q - </span>amount of energy gained or lost.<span> 
</span>
8 0
4 years ago
Read 2 more answers
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.
maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in  of boron trifluoride? moles of fluorine.​

Explanation:

The molecular formula of boron trifluoride is BF_3.

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- 6.023x10^2^3 atoms.

Then, 2.20 moles of boron has

-=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0
3 years ago
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