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3 years ago

Time (sec) [CH4] (mol/L)

Chemistry

1 answer:

Oxana [17]3 years ago

8 0

Based on the table and the equation of reaction; the

rate of consumption of oxygen between time 20 and 40s is 0.005 M/s
rate of consumption of oxygen between time 40 and 60s is 0.0025 M/s

<h3>What is rate of a reaction?</h3>

The rate of a reaction is the rate at which reactants molecules are consumed or products are formed.

Rate of reaction = change in concentration of reactants or products/ time taken

Based on the equation of reaction, the rate of consumption of oxygen is equal to the rate of consumption of methane.

Thus. rate of consumption of oxygen from time 20 seconds to time 60 seconds is calculated as follows:

Rate between 20 and 40 = 1.0 - 0.9/20

Rate of consumption of oxygen = 0.005 M/s

Rate between 40 and 60 = 0.9 - 0.85/20

Rate of consumption of oxygen = 0.0025 M/s

Therefore, the rate of consumption of oxygen decreases with time.

Learn more about rate of reaction at: brainly.com/question/25724512

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3 years ago

Calculate the number of moles equivalent to 12.7 gram of iodine molecule

Alexus [3.1K]

$\LARGE{ \boxed{ \purple{ \rm{Answer}}}}$

☃️ Chemical formulae ➝ $\sf{I_2}$

How to find?

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

$\boxed{ \sf{No. \: of \: moles = \frac{Given \: weight}{Molecular \: weight} }}$

Solution:

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Then, no. of moles,

⇛ No. of moles = 12.7 / 254

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4 years ago

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"Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Alex17521 [72]

Answer:

$\boxed{\text{0.50 mol/L}}$

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

$\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}$

2. Solve for x

$K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}$

3. Calculate the equilibrium concentration of COF₂

c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol

$\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}$

Check:

$\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}$

OK.

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