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RideAnS [48]
3 years ago
9

A certain organ pipe, open at both ends, produces a fundamental frequency of 272 HzHz in air. Part A If the pipe is filled with

helium at the same temperature, what fundamental frequency fHefHef_He will it produce? Take the molar mass of air to be 28.8 g/molg/mol and the molar mass of helium to be 4.00 g/molg/mol .
Chemistry
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

fundamental frequency in helium = 729.8 Hz

Explanation:

Fundamental frequency of an ope tube/pipe = v/2L

where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L  is length of  the pipe

To find the length of the pipe,

frequency  = velocity of sound / 2L

272 = 340 / 2 L

L = 0.625 m

If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.

Also, the velocity of sound is inversely proportional to  square root of molar mass of gas; v ∝ 1/√m

v₁/v₂ = √m₂/m₁

v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium

340 / v = √4 / 28.8

v₂ = 340 / 0. 3727

v₂  =  912.26  m /s  

fundamental frequency in helium  = v₂ / 2L

fundamental frequency in helium = 912.26 / (2 x 0.625)

fundamental frequency in helium = 729.8 Hz

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In a certain experiment, 28.0 mL of 0.250 M HNO3and 53.0 mL of 0.320 M KOH are mixed. Calculate the number of moles of water for
Xelga [282]

Answer:

The number of moles of water formed in the resulting reaction is 6.03

[H+]: 37,2 M

[OH-]: 37,2 M

Explanation:

HNO3  +  KOH ----> KNO3 + H2O

First, we must discover the limiting reagent and we need to find out the moles, we use for this.

Moles that are used = Molarity / volume

HNO3 : 0,250 mol/L / 0,028L = 8,93 moles

KOH : 0,320 mol/L / 0,053L = 6,03 moles

The ratio of the reagents by stoichiometry is 1 to 1, so the limiting reagent is KOH, if I need 1 mole of nitric per mole of KOH, for every 8.93 moles I will need the same. However I have only 6.03 moles of KOH

The ratio of the reagents/products by stoichiometry is 1 to 1 so if I need 1 mol of KOH to make 1 mol of Water, 6,03 moles of KOH are used to make 6,03 moles of H2O.

The equilibrium of water is this:

2H2O ⇄ H3O+  +  OH-

2 moles of water are broken down into 1 mole of hydronium (H3O +) and 1 mole of hydroxyl (OH-)

6,03 moles of water are broken down into the half of those moles, so we have 3,015 moles of H3O+ and 3,015 moles of OH- but these moles are in 81,0 mL (the volume of the two solutions, 28 mL + 53 mL)

We must find out the moles in 1000 mL (1 L) so let's apply the rule of three.

81 mL ____ 3,015 moles

1000 mL ___ ( 1000 . 3,015) /81 = 37,2 M

7 0
2 years ago
Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H
solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol

2) Moles of ammonia gas=\frac{2.53 g}{17 g/mol}=0.1488 mol

Moles of oxygen gas =\frac{3.53 g}{32 g/mol}=0.1101 mol

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

\frac{4}{3}\times 0.1101 mol=0.1468 mol of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

\frac{2}{3}\times 0.1101 mol=0.0734 mol of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100

Percentage yield of the reaction:

\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%

36.37% is the percent yield of the reaction.

3 0
2 years ago
If the ph of a solution is decreased from ph 8 to ph 6 it means that the
saw5 [17]

Explanation:

Relation between pH and concentration of hydrogen ions is as follows.

                  pH = -log [H^{+}]

So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

Therefore, the solution becomes more acidic in nature.

Hence, if the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions has increased in the solution.

5 0
3 years ago
Read 2 more answers
Really need help with this! Chemistry
Westkost [7]

Answer:

a) 0,5

Explanation:

If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5

5 0
2 years ago
22. pH is a measure of ______________.
just olya [345]
22 is a
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24 is c
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6 0
3 years ago
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