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umka21 [38]
3 years ago
10

Energy conversions _____. can lose energy as sound can be 100 percent efficient if controlled properly are calculated by multipl

ying useful energy by the total amount of energy
Chemistry
2 answers:
nikdorinn [45]3 years ago
6 0

Answer:

are calculated by multiplying useful energy by the total amount of energy

Explanation:

The energy conversions are never 100 percent efficient. This is because the energy originally is converted into a mixture of the useful and not useful energy. For example, gasoline in the car is converted to motion, sound and friction. The later two (sound and friction) are energies but are less useful. Thus, in this case, the energy conversion is not 100 % efficient. Thus, the energy efficiency equation comes into play. This is given as the ratio of the useful energy to the total energy. It is a fraction.  

Delicious77 [7]3 years ago
6 0

Answer:

Can lose energy as sound

Explanation:

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Determine the number of atoms in 73.0 grams of calcium, Ca. (The
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# of atoms per mol = Avogadro’s # (6.022 x 10^23)

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2 years ago
How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250
Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are 2.815\times 10^{21}

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{CO_2}

where,

K_H = Henry's constant = 0.034mol/L.atm

C_{CO_2} = molar solubility of carbon dioxide gas

p_{CO_2} = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of carbon dioxide = 8.5\times 10^{-5}M

Volume of solution = 0.550 L

Putting values in above equation, we get:

8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules

So, 4.675\times 10^{-3} moles of carbon dioxide will contain = (6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21} number of molecules

Hence, the number of molecules of carbon dioxide gas are 2.815\times 10^{21}

3 0
3 years ago
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3 years ago
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

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Butane  ⇄  Isobutane

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After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

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