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MissTica
2 years ago
9

How does wavelength affect the pitch of a sound?

Chemistry
1 answer:
boyakko [2]2 years ago
3 0

Answer:

D

Explanation:

because the higher the wavelength the shorter the pitch and the shorter the wavelength the higher the pitch

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Plzz write a small pharghraph
sergiy2304 [10]
Mass = reactencts, 73g
4 0
3 years ago
NO + H2-&gt; N2 + H2O<br> Balance equation?
Oksanka [162]

Answer:

2NO + 4H-> N2 + 2 H2O

Explanation:

Both sides must be equal. :)

5 0
3 years ago
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
What is the correct name for the compound H2O
Deffense [45]

<u>Answer:</u>

The common name for the compound H2O is water.

<u>Explanation:</u>

The systemic name of H2O is Dihydrogen monoxide.

8 0
3 years ago
Read 2 more answers
How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol
Ede4ka [16]

3.74×10^{21}

3.74 ×10^{21} molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane =  0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21×10^{-3} moles of propane

No one mole of propane As a 6.0-2 × 10^{23}

so, 6.21 ×10^{-3} × 6. 022 × 10^23

= 3.74 ×10^{21}

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 ×10^{21}

<h3>What is erlenmeyer flask?</h3>
  • A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
  • It bears the name Emil Erlenmeyer after the German chemist.

<h3>What purpose does an Erlenmeyer flask serve?</h3>
  • Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.
  • For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.

To learn more about calculating total molecules visit:

brainly.com/question/8933381

#SPJ4

4 0
1 year ago
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