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musickatia [10]

2 years ago

9

A certain first-order reaction has a rate constant of 2.65×10−2s−1 at 16 ∘C. What is the value of k at 63 ∘C if Ea = 88.5 kJ/mol

?

1 answer:

alisha [4.7K]2 years ago

5 0

Answer:

chemistry ka mujy b krna☹️

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A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible struct

arsen [322]

Answer:

Plausible structure has been given below

Explanation:

Molar mass of $CO_{2}$ is 44 g/mol and molar mass of $H_{2}O$ is 18 g/mol
Number of mole = (mass/molar mass)

4.04 g of $CO_{2}$ = $\frac{4.04}{44}moles$ $CO_{2}$ = 0.0918 moles of $CO_{2}$

1 mol of $CO_{2}$ contains 1 mol of C atom

So, 0.0918 moles of $CO_{2}$ contains 0.0918 moles of C atom

1.24 g of $H_{2}O$ = $\frac{1.24}{18}moles$ $H_{2}O$ = 0.0689 moles of $H_{2}O$

1 mol of $H_{2}O$ contain 2 moles of H atom

So, 0.0689 moles of $H_{2}O$ contain $(2\times 0.0689)moles$ of $H_{2}O$ or 0.138 moles of $H_{2}O$

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is $C_{2}H_{3}$

So, molecular formula of one of it's analog is $C_{4}H_{6}$

Plausible structure of $C_{4}H_{6}$ has been given below.

4 0

2 years ago

Free brainlest just thank my other questions

alexandr402 [8]

Answer:

ok and thank you for free point

6 0

2 years ago

Read 2 more answers

If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it

Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

$NO_2\rightarrow NO+\frac{1}{2}O_2$

The reaction is second order for $NO_2$ with a rate constant of $0.543M^{-1}s^{-1}$ at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.543M^{-1}s^{-1}$

t = time taken = ?

[A] = concentration of substance after time 't' = 0.150 M

$[A]_o$ = Initial concentration = 0.260 M

Putting values in above equation, we get:

$0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s$

Hence, the time taken is 5.19 seconds

6 0

3 years ago

Explain how a light can be seen when sent through a colloid mixture, but not a solution.

Free_Kalibri [48]

I think the answer is that is lights up?

5 0

2 years ago

A gas cylinder of volume 5.00 l contains 1.00 g of ar and 0.500 g of ne. the temperature is 275 k. find the partial pressure of

amid [387]

<span>11.3 kPa The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = Absolute temperature We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon. Atomic weight argon = 39.948 Atomic weight neon = 20.1797 Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol Now take the ideal gas equation and solve for P, then substitute known values and solve. PV = nRT P = nRT/V P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L P = 113.8892033 L*kPa / 5.00 L P = 22.77784066 kPa Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles. 0.024777375 mol / 0.049809918 mol = 0.497438592 Now multiply by the pressure 0.497438592 * 22.77784066 kPa = 11.33057699 kPa Round the result to 3 significant figures, giving 11.3 kPa</span>

8 0

2 years ago