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yuradex [85]
3 years ago
6

Melting is a physical change in matter, so it doesn’t change the chemical makeup or chemical properties of matter.

Chemistry
1 answer:
stepan [7]3 years ago
3 0

is this site ment for k12


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1H2 + 2O2 -> 3Н2O. ​
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Answer:

1H2 + O2 -> 2Н2O.

combination reaction.

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3 years ago
What information is needed to determine the amount of moles in 1.6g of HCL?
alukav5142 [94]

Answer:

Molar mass and Mass

Explanation:

The relationship between mass and number of moles is given as;

Number of moles = Mass / Molar mass

Mass = 1.6 g

Molar mass of HCl = ( 1 + 35.5 ) = 36.5 g/mol

Number of moles = 1.6 g / 36.5 g/mol

Number of moles = 0.0438 mol

8 0
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Match the layers of the atmosphere
Ira Lisetskai [31]

Answer:

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7 0
3 years ago
Determine the percent composition of phosphorous in calcium phosphate
scoundrel [369]

40% because of ionic charges.

7 0
3 years ago
(a) Calculate the wavelength of light in vacuum that has a frequency of 5.49 ✕ 1018 Hz. 0.0546 Correct: Your answer is correct.
vfiekz [6]

Answer:

a) Wavelength of the light in vacuum = (5.46 × 10⁻¹¹) m = 0.0546 nm

b) Wavelength of the light in diamond = (2.26 × 10⁻¹¹) m = 0.0226 nm

c) Energy of one photon in vacuum = (3.638 × 10⁻¹⁵) J = (2.271 × 10⁴) eV

d) No, the energy of the photon doesn't change when it is travelling inside diamond.

Explanation:

Wavelength (λ), frequency (f) and velocity of light (v) are related as thus

v = fλ

a) v = fλ

v = velocity of light in vacuum = (3.0 × 10⁸) m/s

f = frequency of the light = (5.49 × 10¹⁸) Hz

λ = wavelength of the light = ?

λ = (v/f) = (3.0 × 10⁸) ÷ (5.49 × 10¹⁸)

= (5.46 × 10⁻¹¹) m = 0.0546 nm

b) To find the wavelength of the light in diamond, we need the refractive index of diamond. This is because light, just like all other waves, change their velocities and subsequently their wavelengths in different materials according to the refractive index of the materials.

Refractive index of diamond = 2.42 (from literature)

2.42 = (wavelength of light in vacuum) ÷ (wavelength of light in diamond)

2.42 = 0.0546 ÷ λ

λ = 0.0546 ÷ 2.42 = 0.0226 nm

c) Energy of a photon in vacuum is given as

E = hf

where E = energy in Joules = ?

h = Planck's constant = (6.626 × 10⁻³⁴) J.s

f = frequency of the light in vacuum = (5.49 × 10¹⁸) Hz

E = (6.626 × 10⁻³⁴) × (5.49 × 10¹⁸) = (3.638 × 10⁻¹⁵) J

1 eV = (1.602 × 10⁻¹⁹) J

The amount of the calculated energy in eV

= (3.638 × 10⁻¹⁵) ÷ (1.602 × 10⁻¹⁹) = (2.271 × 10⁴) eV

d) As light travels from material to material, it's velocity and wavelength changes from material to material, but the frequency of the light waves stay the same. Since the energy of the photon depends solely on this frequency, it shows that the energy of the photon stays consistent in whichever material.

Hope this Helps!!!

3 0
3 years ago
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