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mafiozo [28]
3 years ago
15

Find the volume occupied by 18.0 g of helium gas at 29.0 ∘c and 1.33 atm of pressure.

Chemistry
1 answer:
otez555 [7]3 years ago
7 0
We can use the ideal gas law equation to find the volume occupied by the gas
PV = nRT
where 
P - pressure - 1.33 atm x 101 325 Pa/atm = 134 762 Pa
V - volume 
n - number of moles - 18.0 g / 4 g/mol = 4.5 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 29 °C + 273 = 302 K
substituting these values in the equation 
134 762 Pa x V = 4.5 mol x 8.314 Jmol⁻¹K⁻¹ x 302 K 
V = 83.8 L
volume occupied by He is 83.8 L 
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How can you determine which bond in a structure is more polar without using an electronegativity table?
UkoKoshka [18]
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:

I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).
5 0
3 years ago
Read 2 more answers
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
zubka84 [21]

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

6 0
3 years ago
What happens to the grass molecules when it is eaten by the zebra?
Lubov Fominskaja [6]

<u>Answer:</u>

<u>read below</u>

<u>Explanation:</u>

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3 0
3 years ago
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Find the density of an object that has a mass of 5 kg and a<br> volume of 50 cm3.
iren2701 [21]
  • Mass=5kg=5000g
  • Volume=50cm^3

\\ \rm\longmapsto Density=\dfrac{Mass}{Volume}

\\ \rm\longmapsto Density=\dfrac{5000}{50}

\\ \rm\longmapsto Density=100g/cm^3

4 0
3 years ago
I will give the brainliest for who answer this questio but no guessing
xz_007 [3.2K]
IVA: nonmetal/other metals VIIA:halogens VIII: Nobel Gases
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