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3 years ago

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200 grams of an organic sample which contains only carbon, hydrogen and oxygen is analyzed and found to contain 97.30grams of ca

rbon, 16.22grams of hydrogen and the remainder oxygen. what is the empirical formula for the compound?

1 answer:

Klio2033 [76]3 years ago

7 0

The empirical formula is C2H4O

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3 years ago

The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur

vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

$T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K$

$R_1 = \dfrac{1}{7}$

$R_2 = \dfrac{1}{49}$

Thus; $\dfrac{R_2}{R_1} = 7$

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

$In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})$

1.9459 = $\dfrac{Ea}{8.314}* 9.0292 *10^{-5}$

$1.9459*8.314 = Ea * 9.0292*10^{-5}$

$16.1782126= Ea * 9.0292*10^{-5}$

$Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}$

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

$T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K$

$In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})$

$In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})$

$In (\dfrac{R_2}{R_1}) = 0.00357$

$\dfrac{R_2}{R_1}= e^{0.00357}$

$\dfrac{R_2}{R_1}= 1.0035$

where ;

$R_2 = \dfrac{1}7{}$

$R_1 = \dfrac{1}{t}$

Now;

$\dfrac{t}{7}= 1.0035$

t = 7.0245 mins

Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0

3 years ago