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3 years ago

What makes landing the hot air balloon a challenge?

1 answer:

7 0

You have to think about the fact that hot air is constantly being blown up elevating the balloon once that air is taken away it's up to the steering and the wind after that. I know that probably doesn't help did you want to know the science part?

You might be interested in

Why does hot water freeze faster than cold water?

weeeeeb [17]

<h2>Answer:</h2>

<u>Temperature dependency is responsible for the process that hot water freeze faster than cold water.</u>

<h2>Explanation:</h2>

The effect given above is called Mpemba Effect. According to this idea hot water freezes more quickly as compared to cold water. But until now there is no convincing explanation for this strange phenomenon. One idea is that hot containers make better thermal contact with a refrigerator and so conduct heat more efficiently because a good conductor is good fro the transfer of heat. Another idea about this effect is that warm water evaporates more quickly and since this is an endothermic process, it cools the water making it freeze more quickly.

7 0

3 years ago

Read 2 more answers

What statement about gases is true?

kenny6666 [7]

Anything that is true

please write down your question correctly

5 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

3 years ago

Which color in the visible spectrum of hydrogen has the least energy

astraxan [27]

I believe it would be the color Red.

4 0

3 years ago

Jen can take either of two separate roads to drive to work. The first is a toll road that is rarely congested. The second road i

Alchen [17]

<span>A good is rivalrous if its consumption by one consumer prevents or impedes its use by other consumers. Roads are rivalrous to the extent that they are subject to congestion, since congestion is caused by use of the road and impedes the use of the road by others. However, a road which is not subject to congestion, like the toll road in this case, is non-rivalrous. A good is excludable if it is possible to prevent those who have not paid for it from using it. So a toll road is by design an excludable good. Hence, the toll road is rivalrous and excludable.</span>

3 0

3 years ago