Disadvantages of charcoal

Chemistry

1 answer:

ziro4ka [17]3 years ago

6 0

Answer: It is not renewable

Charcoal contains high amount of CO2 emission.

Charcoal when burned emits radiation and even clean coals produces methane.

Explanation:

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An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about

Lisa [10]

Answer:

$C_2H_6$

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

$m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC$

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

$m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O} =0.7molH$

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

$C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3$

Whose molar mass is:

$M_{CH_3}=12+1*3=15g/mol$

Thus, the ratio of the molecular formula to the empirical formula is:

$\frac{30}{15}=2$

Therefore, the molecular formula is twice the empirical formula:

$C_2H_6$

Which is actually ethane.

Regards.

3 0

3 years ago

Vanillin is the compound containing carbon, hydrogen, and oxygen that gives vanilla beans their distinctive flavor. The combusti

kati45 [8]

This website lies, at first you only needed to make a free account to see verified answers and now its sating i need to pay $15 bull
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4 0

4 years ago

An electron in a hydrogen atom drops from energy level e3 to e2. what's the frequency of the emitted photon

castortr0y [4]

For this problem we can use 2 equations.
(1) - E = mC²
E = Energy of photon (J)
m = Mass of photon (kg) (1.67x10⁻²⁷ kg)
C = speed of light (3 x 10⁸ m/s)

(2) - E = hf
E = Energy of photon (J)
h = plank's constant (6.63 × 10⁻³⁴J s)
f = frequency of the photon (Hz)

(1) = (2)
hence, mC² = fh

by rearranging,
f = mC² / h
 f = 1.67x10⁻²⁷ kg * (3 x 10⁸ m/s)² / (6.63 × 10⁻³⁴J s)
 f = 2.27 x 10²³ Hz

3 0

3 years ago

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.