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3 years ago

Hydrocarbons consisting of short carbon chains are ________.

Chemistry

1 answer:

Tresset [83]3 years ago

4 0

Answer:

Explanation:

Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.

As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain

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Answer:

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Explanation:

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2 years ago

Read 2 more answers

At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p

dedylja [7]

Explanation:

The given reaction is as follows.

$2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)$

Value of equilibrium constant is given as $K_{c}$ = 4.3 \times 10^{6}[/tex].

Concentration of given species is $[SO_2]$ = 0.010 M; $[SO_3]$ = 10.M; $[O_2]$ = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

Q = $\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$

Putting the given concentration as follows.

Q = $\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$

Q = $\frac{(10)^{2}}{(0.010)^{2}(0.010)}$

Q = $10^{8}$

It is known that when Q > $K_{eq}$ , then reaction moves in the backward direction.

When Q < $K_{eq}$ , then reaction moves in the forward direction.

When Q = $K_{eq}$ , then reaction is at equilibrium.

As, for the given reaction Q > $K_{eq}$ then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

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3 years ago

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2 years ago

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2 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

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3 years ago