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anzhelika [568]
3 years ago
6

You decide it is time to clean your pool since summer is quickly approaching. Your pool maintenance guide specifies that the chl

orine, Cl2, concentration of the pool should be between 1 and 3 ppm. In order to determine if your pool is safe to swim in, you send a sample of pool water to a chemist for analysis of the Cl2 content. The chemist reports a chlorine concentration of 2.96 × 10–5 M. Convert the concentration of Cl2 to parts-per-million (ppm).
Physics
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

Concentration of Cl2: 1,048 ppm

Explanation:

The unit of measurement molarity (M) represents the same magnitud as mole/L (number of moles of the substance in one liter of solution). The concentration in ppm represents the same as mg/L (number of miligrammes of the substance in one liter of solution). In order to convert from M to ppm we have to consider, then, mass and volume. We can see that in both cases the volume is expressed in liters, so we don't have to do anything to change it. The problem comes when you see that you have to convert moles, from the molatiry, to miligrammes, to get the result in mg/L, meaning ppm.

First of all, notice that the substance is Cl2, so we need to find a relationship between the number of moles and its mass. If you look in the periodic table you'll see that the atomic mass for Cl is 35,4 g/mole (grammes of Cl in one mole). So, there is a way to relate the moles to the mass of the substance and it is represented on the equation below:

mass=mole*atomic mass

We want to find the mass (m) and we know the amount of moles of Cl2 in the solution (moles=2,96x10^-5), so, if we use the values known on the equation above we get that:

m=2,96x10^-5 moles*35,4 g/mole\\ m=1,048x10^-3 g

Remember that these grammes are found in one liter of solution. So, this means we have 1,048x10^-3 g/L. Previously we said that ppm=mg/L, so all that's left to do it to convert grammes to miligrammes:

1 g = 1000 mg

If we multiply both sides by 1,048x10^-3:

1 * 1,048x10^-3 g = 1000 * 1,048x10^-3 mg

1,048x10^-3 g = 1,048 mg

Knowing that this amount of mass was found in one liter, we get that the amount of substance in the solution is:

1,048 mg/L

Knowing that <em>mg/L=ppm</em>, then the concentration of Cl2 in ppm is:

<u>1,048 ppm</u>

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An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
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The weight of the object is   

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                   Weight = (60 kg) x (9.8 m/s²)

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Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

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___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
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myrzilka [38]

Answer:

(a) 0

(b) 10ML

(c) 10ML(1 - cos(\theta))

(d) 10ML(1 + sin(\phi))

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

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(c) At angle \theta from the vertical. Vertically speaking, the child should be at a distance of Lcos(\theta) to the swinging point, and a vertical distance of L - Lcos(\theta) to the lowest position. His potential energy to this point would be:

E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))

(d) at angle \phi from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of Lsin(\phi) from the swinging point and higher than it. Therefore his vertical distance to the lowest point is L + Lsin(\phi) = L(1 + sin(\phi))

His potential energy to his point would be:

E_{\phi} = mgh = 10ML(1 + sin(\phi))

5 0
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