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3 years ago

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Have your mother and father (or any adult man and woman) sing together on the same note, at the same pitch (not an octave apart)

. Write down any differences you can hear. Is one voice lower or higher sounding (even though they are singing the same pitch)? Are there other differences you can hear? Explain your results using the terms “overtones,” “frequency,” “pitch,” and “sound quality.”

1 answer:

wlad13 [49]3 years ago

5 0

Answer:

No one sounds the same in singing everyone has a different pitch, overtones, and sound quality. Now in some songs people harmonize together and that can sound the same in pitch but there are still different pitches and tones.

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a sample of gas initially occupies 5.50 liter at a pressure of 0.750 atm at 13.0 C. what will the temperture change to 22.5 C an

sergey [27]

Answer:

The pressure will be 2.84 atm

Explanation:

The gas laws are a set of chemical and physical laws that allow us to determine the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.

Boyle's law indicates that the volume is inversely proportional to the pressure: if the pressure increases, the volume decreases and if the pressure decreases, the volume increases. Boyle's law is expressed mathematically as:

P * V = k

On the other hand, Charles's law indicates that with increasing temperature, the volume of the gas increases and with decreasing temperature, the volume of the gas decreases. That is, they are directly proportional. This is expressed mathematically as:

$\frac{V}{T} =k$

Finally, the Gay-Lussac law says that at constant volume, the pressure of the gas is directly proportional to its temperature. Mathematically, it is expressed as:

$\frac{P}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T}=k$

When you want to study two different states, an initial and a final one of a gas, you can use the following expression:

$\frac{P1*V1}{T1}=\frac{P2*V2}{T2}$

In this case:

P1= 0.750 atm
V1= 5.50 L
T1= 13 C= 286 K (being 0 C=273 K)
P2=?
V2= 1.50 L
T2= 22.5 C=295.5 K

Replacing:

$\frac{0.750 atm*5.50 L}{286 K}= \frac{P2*1.50 L}{295.5 K}$

Solving:

$P2=\frac{295.5K}{1.50L} *\frac{0.750 atm*5.50 L}{286 K}$

P2= 2.84 atm

<u><em>The pressure will be 2.84 atm</em></u>

8 0

4 years ago

Enter the formula for the polyatomic ion in MnCO3.

joja [24]

The formula is Mn2+CO2= MnCO3

8 0

3 years ago

4. Comment les géologues ont-ils déterminé les limites entre les plaques tectoniques ?

Alex787 [66]

Answer:

i do not know

Explanation:

4 0

3 years ago

11. Assuming that the gases are ideal, calculate the amount of work done (joules) in each of the following reactions at 25 degre

gayaneshka [121]

<u>Answer:</u>

<u>For a:</u> Work done for the given reaction is 2477.572 J.

<u>For b:</u> Work done for the given reaction is 0 J

<u>Explanation:</u>

To calculate the work done for the reaction, we use the equation:

$W=-P\Delta V$

Ideal gas equation follows:

$PV=nRT$

Relating both the above equations, we get:

$W=-\Delta n_gRT$ ......(1)

where,

$\Delta n_g$ = difference in number of moles of products and reactants = $n_g_{(products)}-n_g_{(reactants)}$

R = Gas constant = 8.314 J/K.mol

T = temperature = $25^oC=[273+25]K=298K$

<u>For a:</u>

The chemical reaction follows:

$4HCl(g)+O_2(g)\rightarrow 2Cl_2(g)+2H_2O(g)$

$\Delta n_g=4-5=-1$

Putting values in equation 1, we get:

$W=-(-1mol)\times (8.314J/K.mol)\times 298K=2477.572J$

Hence, work done for the given reaction is 2477.572 J.

<u>For b:</u>

The chemical reaction follows:

$2NO(g)\rightarrow N_2(g)+O_2(g)$

$\Delta n_g=2-2=0$

Putting values in equation 1, we get:

$W=-(0mol)\times (8.314J/K.mol)\times 298K=0J$

Hence, work done for the given reaction is 0 J.

3 0

3 years ago

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is burned in a bomb calorim

Serggg [28]

Answer:

$\Delta E_{rxn}$ for combustion of naphthalene is -5164 kJ/mol

Explanation:

$\Delta E_{rxn}=C_{calorimeter}\times \Delta T_{calorimeter}$

where, C refers heat capacity and $\Delta T$ refers change in temperature.

Here, $\Delta T_{calorimeter}=(32.33-24.25)^{0}\textrm{C}=8.08^{0}\textrm{C}$

So, $\Delta E_{rxn}=(5.11\frac{kJ}{^{0}\textrm{C}})\times (8.08^{0}\textrm{C})=41.3kJ$

$\Delta E_{rxn}$ is generally expressed in terms of per mole unit of reactant. Also, $\Delta E_{rxn}$ should be negative as it is an exothermic reaction (temperature increases).

Molar mass of naphthalene is 128.17 g/mol

So, 1.025 g of naphthalene = $\frac{1.025}{128.17}moles$ of naphthalene

= 0.007997 moles of naphthalene

$\Delta E_{rxn}=-\frac{41.3kJ}{0.007997mol}=-5164kJ/mol$

5 0

3 years ago