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2 years ago

Rank the following from lowest to highest anticipated boiling point: C8H18, CH, C16H34,CH4,C4H10

1 answer:

8 0

The relative strength of intermolecular forces such as ionic, hydrogen bonding, dipole-dipole interaction and Vander Waals dispersion force affects the boiling point of a compound. For this case, the longer the chain the higher the boiling point.

CH, CH4, C4H10, C8H18, C16H34

Hope this answers the question. Have a nice day.

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What cell part is NOT found in all cells?

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heart I think also I'm so sorry if I get you this wrong I think membrane not sure tell me what was the right answer but I think membrane.

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2 years ago

The temperature of a sample of water increases from 20c to 46.6c as it absorbs 5650 Joules of heat. What is the mass of the samp

WINSTONCH [101]

Answer: 51 g

Explanation:

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1 year ago

Which unit represents 100 meters?

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Answer:

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2 years ago

Read 2 more answers

a solution of sodium hydroxide, 0.0500 M, is used to titrate a 15.00 mL sample of hydrochloric acid to the endpoint. The initial

Akimi4 [234]

Answer:

The answer to your question is Molarity = 0.0708

Explanation:

Data

NaOH 0.05 M Volume 1 = 3.87 ml Volume 2 = 25.11 ml

HCl 15 ml

Process

1.- Find the volume used of NaOH

25.11 - 3.87 = 21.24 ml = 0.02124 l

2.- Write the balanced equation of the reaction

NaOH + HCl ⇒ NaCl + H₂O

3.- Calculate the moles of NaOH in the solution

Molarity = $\frac{moles}{volume}$

moles = Molarity x volume

moles = 0.05 x 0.02124

moles = 0.001062

4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.

1 mol of NaOH ------------- 1 mol of HCl

0.001062 moles of NaOH ------------ x

x = (0.001062 x 1) / 1

x = 0.001062 moles of HCl

5.- Find the molarity of HCl

Molarity = $\frac{0.001062}{0.015}$

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4 0

2 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

Where:

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

3 years ago