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maksim [4K]
3 years ago
12

Which formula represents an unsaturated hydrocarbon? (1) C5H12 (3) C7H16 (2) C6H14 (4) C8H14

Chemistry
2 answers:
Alik [6]3 years ago
7 0

Answer:

Answer is C₈H₁₄

Explanation:

The general molecular formula of alkanes (saturated hydrocarbons) is:

CₙH₂ₙ₊₂

Where n = number of carbon atoms.

The general molecular formula of alkenes (unsaturated hydrocarbons) is:

CₙH₂ₙ

The general molecular formula of alkynes (unsaturated hydrocarbons) is:

CₙH₂ₙ₋₂

Now let us check the formula of given compounds

a)C₅H₁₂

n = 5

2n=10

2n+2=12

so this is an alkane, saturated hydrocarbon.

b)C₇H₁₆

n = 7

2n=14

2n+2=14

so this is an alkane, saturated hydrocarbon.

c)C₆H₁₄

n = 6

2n=12

2n+2=14

so this is an alkane, saturated hydrocarbon.

d) C₈H₁₆

n = 8

2n=16

2n+2=18

2n-2=14

so this is an alkyne, unsaturated hydrocarbon.

Delicious77 [7]3 years ago
5 0
(4) C8H14 is your answer. A saturated hydrocarbon means that they have no double or triple bonds. C8H14 is an alkyne, which is an unsaturated hydrocarbon.
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Calculate the standard cell potential of the following cell at 25°C. Sn(s) | Sn2+(aq) || Cu2+(aq) | Cu(s).
Semmy [17]

Answer:

0.48 V

Explanation:

Usually in the cell notation, the left side shows oxidation. So,

Oxidation half reaction:

Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-    E^o_{ox}=-E^o_{red}=-(-0.14\ V)=0.14\ V

Reduction half reaction:

Cu^{2+}_{(aq)}+2e^-\rightarrow Cu_{(s)}    E^o_{red}=0.34\ V

E^o_{cell}=E^o_{ox}+E^o_{red}= (0.14+0.34)\ V=0.48\ V

4 0
3 years ago
A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s
marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

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