If one molecule of glucose (C6H120) reacts completely with three molecules of oxygen (O2), what is the TOTAL number of atoms

Chemistry

1 answer:

nalin [4]2 years ago

5 0

Answer : The total number of atoms in the product are, 36 atoms.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When 1 molecule of glucose reacts completely with 3 molecules of oxygen then it react to give 6 molecules of carbon dioxide and 6 molecules of water.

The balanced chemical reaction will be:

$C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O$

There are 3 atoms in carbon dioxide and 3 atoms in water.

Total number of atoms in the product = 6(3) + 6(3) = 36 atoms

Thus, the total number of atoms in the product are, 36 atoms.

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

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