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2 years ago

If a helium balloon is put into a hot oven. What happens to the volume of the balloon?

Chemistry

1 answer:

AveGali [126]2 years ago

6 0

Answer:

What about hot air balloons? They work by similar principles. If you heat up a gas it expands. In the case of a hot air balloon, when the gas inside the balloon expands the extra gas is pushed out the bottom of the balloon, meaning that there are fewer atoms inside the balloon, meaning that the air in the balloon is lighter than the air outside the balloon.

The amount of lifting power is controlled by how hot the air is. If you heat the air inside the balloon 100 degrees F hotter than the outside air temperature, then the air inside the balloon will be about 25 percent lighter than the air outside the balloon. So a cubic foot of air weighs about 35 grams at 32 degrees F. A cubic foot of hot air at 132 degrees F will weigh 25 percent less, or about 26.5 grams. The difference is 8.5 grams or so. So a hot air balloon has to be much bigger to support the same weight, but it will float because hotter air is lighter than cooler air.

Explanation:

The amount of lifting power

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Nady [450]

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA $\rightarrow$ citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a). It is known that , relation between change in free energy ( $\Delta G$ ) of a reaction and equilibrium constant (K) is as follows.

$\Delta G = -RT \times ln K$

where, T = temperature in Kelvin

The given data is as follows.

T = 310 K, $\Delta G = -31.5 kJ /mol = -31500 J/mol$ (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

ln K = $\frac{-(\Delta G)}{RT}$

= $\frac{31500}{8.314 \times 310}$

ln K = 12.22

K = antilog (12.22)

= $2.1 \times 10^{5}$

Therefore, we can conclude that value of equilibrium constant for the given reaction is $2.1 \times 10^{5}$ .

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2 years ago

What is the final volume of NaOH solution prepared from 100.0 mL of 0.500 M NaOH if you wanted the final concentration to be 0.1

suter [353]

Answer:

The final volume of NaOH solution is 30ml

Explanation:

We all know that

V1S1 = V2S2

or V1= V2S2÷S1

or V1= V2×S2×1/S1

or V1=100×0.15×1/0.50

V1= 30

∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.

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3 years ago

Haw many H atom in tow H2O molecules

allochka39001 [22]

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2 years ago

A hypothetical element, E, has two stable isotopes. One isotope has a natural abundance of 68.037% and has an atomic mass of 46.

I am Lyosha [343]

If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.

Substituting into the atomic mass formula,

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1 year ago

2. 2. Consider the following reaction occurring in a closed chemical system. Assume that this reaction is at equilibrium and tha

Alika [10]

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3 years ago

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