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3 years ago

How are scented candles made?

Chemistry

2 answers:

marysya [2.9K]3 years ago

7 0

Answer:

The aroma from a lighted scented candle is released through the evaporation of the fragrance from the hot wax pool and from the solid candle itself. Like unscented candles, properly-formulated scented candles will primarily produce water vapor and carbon dioxide when burned

Explanation:

Gnom [1K]3 years ago

5 0

Like unscented candles, properly-formulated scented candles will primarily produce water vapor and carbon dioxide when burned. ... Most scented candles contain a combination of natural and synthetic fragrances These fragrance materials may be derived from essential oils or from synthetic aroma chemicals.
G O O G L E :)

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PLEASE HELP!! WILL MARK BRAINLIEST AND THANK YOU!! EXTRA POINTS!!

babunello [35]

Answer:

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3 years ago

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Len [333]

Answer:

Explanation:

There are 4 sulfur atoms in SO4

4×3=12

This means that it turns into 3×(SO4)

=3SO4

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3 years ago

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Norma-Jean [14]

Answer:

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3 years ago

Read 2 more answers

For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of

torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For $Al_2O_3$

Mass of $Al_2O_3$ = 4.07 g

Molar mass of $Al_2O_3$ = 101.96 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.07\ g}{101.96\ g/mol}$

$Moles\ of\ Al_2O_3= 0.0399\ mol$

According to the reaction:

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

1 mole of $Al_2O_3$ on reaction produces 1 mole of $Al_2(SO_4)_3$

So,

0.0399 mole of $Al_2O_3$ on reaction produces 0.0399 mole of $Al_2(SO_4)_3$

Moles of $Al_2(SO_4)_3$ obtained = 0.0399 mole

Molar mass of $Al_2(SO_4)_3$ = 342.2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0399= \frac{Mass}{342.2\ g/mol}$

$Mass= 13.7\ g$

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

$\%\ yield =\frac{10.4}{13.7}\times 100$

<u>% yield =76 %</u>

6 0

3 years ago

Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________at

MatroZZZ [7]

Answer:

D. 1.48atm

Explanation:

Van der waals equation is given as:

(P +an²/v²) (v - nb) = nRT

Where;

P = pressure (atm)

V = volume (L)

R = gas constant (0.0821 Latm/molK)

a and b = gas constant specific to each gas

T = temperature (K)

n = number of moles

According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol

Hence;

(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15

(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638

(P + 0.0291) (22.316) = 33.638

22.316P + 0.649 = 33.638

22.316P = 33.638 - 0.649

22.316P = 32.989

P = 32.989/22.316

P = 1.478

P = 1.48atm

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3 years ago