hydrogen gas and oxygen gas react to form liquid hydrogen peroxide. which starement is true about the reaction?

Chemistry

1 answer:

Strike441 [17]3 years ago

4 0

The two elements mixed together to form a chemical bond and reacted as a chemical change

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What takes place during a redox reaction

VladimirAG [237]

Electrons are both gained and lost.

7 0

3 years ago

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Commercial cold packs often contain solid NH4NO3 and a pouch of water. The temperature of the pack drops as the NH4NO3 dissolves

Yanka [14]

Answer: b. $\Delta H_{soln}$ is positive and $\Delta S_{soln}$ is positive.

Explanation:-

As the temperature of the pack drops, the energy has been absorbed from the pack for dissolution of $NH_4NO_3$ in water. Thus as the energy has been absorbed in the reaction, the reaction is endothermic and the change in enthalpy i.e. $\Delta H_{soln}$ is positive.

The entropy is the measure of degree of randomness. The entropy increases when the randomness increases and the entropy decreases when the randomness decreases. When a substance dissolves in water, it dissociate into ions and hence the randomness increases thus the change in entropy i.e. $\Delta S_{soln}$ is positive.

$NH_4NO_3\rightarrow NH_4^++NO_3^-$

7 0

3 years ago

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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