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2 years ago

24 What is the atomic number of the element whose atoms bond to each other in chains, rings, and networks?(1) 10 (3) 6

Chemistry

2 answers:

Marysya12 [62]2 years ago

4 0

Answer: The correct answer is option (3).

Explanation:

The property of the atoms of an element to bind with each other in form chains,rings and networks is known as Catenation. Generally, this property is widely shown by the carbon atom , This is the reason that vast number of organic molecules are made up carbon atoms.

The atomic number of carbon atom is 6. Hence the correct answer is option(3).

strojnjashka [21]2 years ago

3 0

It's carbon, which has an atomic number of 6.

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The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

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$T_2$ = final temperature =383 K

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Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

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