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3 years ago

The following are in order of increasing boiling temperatures: (A) RbCl

Chemistry

1 answer:

andrew11 [14]3 years ago

5 0

Answer:

Option E

RbCL<CH3OH< CH3Cl< CH4

Explanation:

The boiling temperature of an organic compound increases with the increase in number of carbon atoms and hence as the length of the hydrocarbon chain increases the boiling temperature also increase due to increase in force of attraction between the molecules. Also presence of strong molecules such as OH – COOH also increases the boiling temperature while Cl and other elements with high electronegativity reduces the boiling temperature

Hence, the correct order of increasing boiling temperature is option E

RbCL<CH3OH< CH3Cl< CH4

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Over a period of 30 days a student studied the night sky and recorded his observations. One night he watched a shooting star str

Temka [501]

Answer:

meteor

Explanation:

The most logical identification of a "shooting star" is a meteor. A meteor is basically any material from outerspace that falls to Earth. The main characteristic of a meteor is that from our point of view on the surface of the Earth it looks like a shooting star because we see a streak of light behind it. This light is simply dust and rock from the meteor burning up as it enters the Earth's atmosphere, leaving a "tail" of light behind it. Which to us looks like a shooting star.

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3 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

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3 years ago

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