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Step2247 [10]
3 years ago
8

The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be

lowered 3.00 cm. If water is pumped out at a rate of 3.80 liters per second, how long will it take to lower the water level 3.00 cm?
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Initial dimension of the pool:

Length (L)= 50.0 m

Width (W)= 25.0 m

Depth (D) = 2.0 m

Initial Volume of the pool (V1)  = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3

When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m

The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3

Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3

Now, 1 m3 = 1000 L

therefore, 35.5 m3 == 35.5 *10^3 L

It is given that:

3.80 L of water is pumped out in 1 sec

Therefore, 35.5*10^3 L will be pumped out in:

= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec



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A discharging lead-acid battery is best described as
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2 years ago
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A sample of iron metal is placed in a graduated cylinder. it is noted that 10.4 ml of water is displaced by the iron. the iron i
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<h3>Answer:</h3>

                 162.43 g of FeCl₂

<h3>Explanation:</h3>

Step 1: Calculate mass of Fe;

As,

                                   Density  =  Mass ÷ Volume

Or,

                                   Mass  =  Density × Volume

Where Volume is the volume of water displaced  =  10.4 mL

Putting values,

                                   Mass  =  7.86 g.mL⁻¹ × 10.4 mL

                                   Mass  =  81.744 g of Fe

Step 2: Calculate amount of FeCl₂;

The balance chemical equation is as follow,

                                Fe  +  2 HCl   →    FeCl₂  +  H₂ ↑

According to this equation,

       55.85 g (1 mol) Fe produced  =  110.98 g (1 mol) of FeCl₂

So,

               81.744 g Fe will produce  =  X g of FeCl₂

Solving for X,

                    X  =  (81.744 g × 110.98 g) ÷ 55.85 g

                     X =  162.43 g of FeCl₂

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The correct answer is b

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