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erica [24]
3 years ago
11

A sound with more vibrations per second sounds higher than a sound with fewer vibrations per second.

Chemistry
2 answers:
Fudgin [204]3 years ago
6 0

Answer:

that is a false sentence

kondor19780726 [428]3 years ago
6 0

Answer:

I'm sure you've already finished the est but the real answer is true, I took the k12 test and got that one correct, hope this helps anyone!

Explanation:

You might be interested in
What is the freezing point of an aqueous solution that has 25.00 g of calcium iodide dissolved in 1250 g of water?
ozzi

Answer:

<u></u>

  • <u>- 0.380ºC</u>

Explanation:

The lowering of the freezing point of a solvent is a colligative property ruled by the formula:

  • \Delta T_f=K_f\times m\times i

Where:

  • ΔTf is the lowering of the freezing point
  • Kf is the molal freezing constant of the solvent: 1.86 °C/m
  • m is the molality of the solution
  • i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.

<u />

<u>a) molality, m</u>

  • m = number of moles of solute/ kg of solvent
  • number of moles of CaI₂ = mass in grams/ molar mass
  • number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
  • m = 0.0850667mol/1.25 kg = 0.068053m

<u>b) i</u>

  • Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3

<u />

<u>c) Freezing point lowering</u>

  • ΔTf =  1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC

<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
Download pdf
6 0
3 years ago
THIS IS URGENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Pani-rosa [81]

Answer:

1- 1.54 mol.

2- 271.9 kPa.

3- Yes, the tires will burst.

4- 235.67 kPa.

5- As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in image 1 that represents the Maxwell’s Distribution of Speeds of molecules. "Kindly, see the explanation and the attached images".

<em>Explanation:</em>

<em>Q1- How many moles of nitrogen gas are in each tire?  </em>

  • To calculate the no. of moles of nitrogen gas in each tire, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the nitrogen gas (P = 247.0 kPa/101.325 = 2.44 atm),

V is the volume of the nitrogen gas (V = 15.2 L),

n is the no. of moles of the nitrogen gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the nitrogen gas (T = 21°C + 273 = 294 K).

∴ n = PV/RT = (2.44 atm)(15.2 L)/(0.082 L/atm/mol.K)(294.0 K) = 1.54 mol.

<em>Q2: What would the maximum tire pressure be at 50 degrees C?  </em>

  • Now, the temperature is raised to be 50°C (T = 50°C + 273 = 323 K).
  • The pressure can be calculated using the general gas law: PV = nRT.

<em>∴ P = nRT/V </em>= (1.54 atm)(0.082 L/atm/mol.K)(323.0 K)/(15.2 L) = 2.68 atm = <em>271.9 kPa.</em>

<em>Q3: Will the tires burst in Moses Lake? Explain.</em>

  • <em>Yes,</em> the tires will burst because the internal pressure be 271.9 kPa that exceeds 270 kPa, the pressure above which the tires will burst.

<em>Q4: If you must let nitrogen gas out of the tire before you go, to what pressure must you reduce the tires before you start your trip? (Assume no significant change in tire volume.)  </em>

  • To get the pressure that we must begin with:
  • Firstly, we should calculate the no. of moles at:

T = 55°C + 273 = 328 K,

Pressure = 270 kPa (the pressure above which the tires will burst). (P =270 kPa/101.325 = 2.66 atm).

V = 15.2 L, as there is no significant change in tire volume.

∴ n = PV/RT = (2.66 atm)(15.2 L)/(0.082 L.atm/mol.K)(328 K) = 1.5 mol.

  • 1.5562 moles of N₂ in the tires will give a pressure of 270 kPa at 55°C, so this is the minimum moles of N₂ that will make the tires burst.
  • Now, we can enter this number of moles into the original starting conditions to tell us what pressure the tires will be at if we start with this number of moles of N₂.

P = ???  

V = 15.6 L.

n = 1.5 mol

T = 21°C + 273 = 294.0 K  

R = 0.0821 L.atm/mol.K.

∴ P = nRT/V = (1.5 mol x 0.082 x 294.0 K) / (15.6 L) = 2.2325 atm = 235.67 kPa.

<em>So, the starting pressure needs to be 235.67 kPa or just under in order for the tires not to burst.</em>

<em />

<em>Q5: Create a drawing of the tire and show a molecular view of the air molecules in the tire at 247 kpa vs the molecular view of the air molecules after the tires have been heated. Be mindful of the number of molecules that you use in your drawing in the before and after scenarios. Use a caption to describe the average kinetic energy of the molecules in both scenarios.</em>

<em />

  • As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in “image 1” that represents the Maxwell’s Distribution of Speeds of molecules.
  • The no. of molecules that possess a critical K.E. of molecules increases due to increasing the temperature activate the motion of molecules with high velocity as
  • (K.E. = 3RT/2), K.E. directly proportional to the temperature of the molecules (see image 2).
  • Also, the average speed of molecules increases as the K.E of the molecules increases (see image 3).

3 0
2 years ago
Provide at least three indicators of a chemical change that are present in burning a piece of paper.
Pie

Answer:

Some chemical indicators perceived while a piece of paper is burning are:

Production of an Odor: there is a smell of burnt paper

Change in Temperature:  combustion is a highly exothermic reaction , so the temperature increase

Change in Color: paper changes to ashes as the burning process occurs

3 0
3 years ago
A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr
Oksanka [162]

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = t_{1/2}=25 minutes

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{25 min}=0.02772 min^{-1}

Initial concentration of A = [A]_o=0.900 M

Final concentration of A after 50 minutes = [A]=?

t = 50 minute

[A]=[A]_o\times e^{-kt}

[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

7 0
3 years ago
Based on the data given, which factor had the greatest effect on the behavior of the gas
Pepsi [2]
Do you have the picture of the data?
6 0
3 years ago
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