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3 years ago

Why does a candle relight if a match is held above the wick just after the flame is blown out?

Chemistry

1 answer:

shutvik [7]3 years ago

5 0

Answer with Explanation:

A candle relights when a match is held above the wick because its trail of smoke still contains some of the wax. When candles are burned, the heat of the flame turns the the wax (which is originally solid) into liquid (commonly near the wick) and then evaporates as gas. The vaporized wax actually protect the wick and this is the reason why it is not burned. So, when you put off a candle, the vaporized wax is still present near the wick. This, remember, holds heat and light energy. Thus, this explains why the candle can be relighted once you hold a match above the wick. It then allows the match to ignite.

Thus, this explains the answer.

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Select the correct answer. Which equation, when solved, gives 8 for the value of x?

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4 years ago

What is the term for a process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concent

Lubov Fominskaja [6]

The answer is:
Osmosis

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3 years ago

Products: 1-methylcyclohexene, 3-methylcyclohexene, methylenecyclohexane

Lostsunrise [7]

Answer:

See explanation

Explanation:

The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.

Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.

8 0

3 years ago

1. The emission spectrum of mercury atoms has a bright green line with wavelength

SashulF [63]

Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.

From the formula;

E = hc/λ

h = Plank's constant = $6.6 * 10^-34$ Js

c = speed of light= $3 * 10^8$

λ = wavelength = $546.1 * 10^-9$ m

E = $6.6 * 10^-34 * 3 * 10^8/546.1 * 10^-9$

E = $3.63 * 10^-19$ J

Also;

E =hf

Where;

h = Planks's constant

f = frequency of photon

f = E/h

f = $3.63 * 10^-19 J/6.6 * 10^-34$

f = $5.5 * 10^14$ Hz

Learn more: brainly.com/question/18415575

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3 years ago

Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2

ArbitrLikvidat [17]

Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>

6 0

3 years ago

Read 2 more answers