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Lera25 [3.4K]
3 years ago
8

A plane has a cruising speed of 100 miles per hour when there is no wind. At this speed, the plane flew 600 miles with the wind

in the same amount of time it flew 400 miles against the wind. find the speed of the wind
Chemistry
1 answer:
alina1380 [7]3 years ago
8 0
The solution for this problem is:
Let x = speed of wind

Speed of plane with the wind = x + 100

Speed of plane against the wind = 100 -x 

 We will be using the formula for distance which is (Rate)(Time), getting the formula for time would be distance/rate Time to travel 600 miles with the wind = Time to travel 400 miles against the wind 600/(x + 100) = 400/(100 - x)

 400(x + 100) = 600(100 - x)

 400x + 40000 = 60000 - 600x

 1000x = 20000

x = 20000/1000

 x = 20 mph
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It is not a gas because its particles do not have large spaces between them.

Explanation:

Solids and liquids have a lot of particles with few spaces between them. Also, gas particles move rapidly in each direction.

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3 years ago
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
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Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

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3 0
3 years ago
Read 2 more answers
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Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

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Notice, we had to convert L to mL to cancel units.

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We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

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We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

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2 years ago
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Answer:

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From the question we have

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We have the final answer as

<h3>59.1 m</h3>

Hope this helps you

3 0
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