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2 years ago

A compound is 150.86 g carbon, 8.86 g hydrogen, and 20.10 g oxygen by

Chemistry

1 answer:

Ivahew [28]2 years ago

8 0

Answer:

C20 H14 O2

Explanation:

Remark

This is a sample, which the question does not say and should. It is a fraction of 1 mole. So what you have to do is multiply the numbers given by x and equate it to 286.28

Equation

150,86* x + 8.86*x + 20.1*x = 286.28

179.8x = 286.28

x = 286.26/179.8

x = 1.592

Now multiply the given numbers by 1.592

150.86 * 1.592 = 240.58

8.85 * 1.592 = 14.1

20.1 * 1.592 = 32

Rounding you get

240/12 = 20

14.1/1 = 14

32/16 = 2

C20 H14 O2

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Suppose the flask is already at equilibrium but then the volume of the reaction flask is then reduced what will happen

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1) Answer is: c) The reaction will proceed right.

Balanced chemical reaction: N₂(g) + 3H₂(g) ⇄ 2NH₃(g) ΔH = +92 kJ.

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3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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