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maria [59]
3 years ago
8

Chicago is about 2,800 km from Los Angeles. About how long would it take a jet plane flying 800 km/h to fly from Chicago to Los

Angeles? A. 2.0 h B. 2.5 h C. 3.0 h
Chemistry
1 answer:
BaLLatris [955]3 years ago
4 0

<u>Answer:</u> The time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

<u>Explanation:</u>

Speed is defined as the rate at which an object moves with respect to time.

To calculate the time taken for the given speed, we use the equation:

s=\frac{d}{t}

where,

s = speed of the jet plane = 800 km/hr

d = distance traveled = 2800 km

t = time taken by jet plane = ?

Putting values in above equation, we get:

800km/hr=\frac{2800km}{t}\\\\t=\frac{2800km}{800km/hr}=3.5hr

Hence, the time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

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when  heat gained = heat lost 

when AL is lost heat and water gain heat

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when M(Al) is the mass of Al= 225g 

C(Al) is the specific heat of Al = 0.9 

ΔT(Al) = (125.5 - Tf) 

and Mw is mass of water  = 500g

Cw is the specific heat of water = 4.81 

ΔT = (Tf - 22.5) 

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

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Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur
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A bowling ball and basketball collide, what direction do they go and how is energy transferred?
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An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

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3 years ago
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