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3 years ago

Describe a chemical formula give an example

1 answer:

3 0

A chemical formula identifies each constituent element by its chemical symbol and indicates the proportionate number of atoms of each element.

<em>For example, the empirical formula of ethanol may be written C2H6O because the molecules of ethanol all contain two carbon atoms, six hydrogen atoms, and one oxygen atom.</em>

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If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is

Aliun [14]

Answer:

486 nm

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

Next, we shall determine the frequency of the photon. This can be obtained as follow:

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

7 0

3 years ago

What is the study of his matter and energy interact? A. Chemistry B. Physics C. Biology D. Planetary science

e-lub [12.9K]

The answer is option (B) Physics

8 0

3 years ago

Read 2 more answers

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$