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konstantin123
2 months 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:
Yuliya22 [3.3K]2 months ago
4 0

Answer:

Cl2 concentration: 1,048 ppm

Explanation:

Molarity (M) is a unit that indicates how many moles of a substance are in one liter of solution, while ppm is similar to mg/L, representing the mass of the substance per liter. To convert M to ppm, we need to consider both mass and volume. The volume remains in liters in both cases, so no adjustments are needed there. The challenge arises when converting moles from molarity to milligrams for the mg/L conversion to find ppm.

First, we must recognize that the substance in question is Cl2, prompting the need to connect mole quantity to its mass. Referring to the periodic table, we find the atomic weight of Cl is 35.4 g/mole. This relationship allows us to establish moles' mass, as shown in the following equation:

mass=mole*atomic mass

We already know the moles of Cl2 in the solution is (moles=2.96x10^-5). Substituting this value, we can calculate:

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

Keep in mind that we're discussing the mass found per liter of solution, which gives us 1.048x10^-3 g/L. We previously established that ppm equals mg/L, so we need to convert grams to milligrams:

1 g = 1000 mg

Now multiplying both sides by 1.048x10^-3:

1 * 1.048x10^-3 g = 1000 * 1.048x10^-3 mg

Thus, 1,048x10^-3 g translates to 1,048 mg.

This amount of mass detected in one liter indicates that the concentration of the substance in the solution is:

1,048 mg/L

Since we know that mg/L=ppm, the conclusion is:

1,048 ppm

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
kicyunya [3294]

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let V_{R}represent the Voltage across the Resistor

V_{c}and

represent the Voltage across the capacitor

Implementing KVL;

emf - V_{R} - V_{c} = 0\\

.........................(1)

Since this is a series connection, the same current traverses through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Integrating V_{c}and V_{R}into equation (1)

emf - IR - Q/C = 0

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Substituting the values of t and I₀ into the aforementioned formula for I

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

4 0
19 days ago
A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is
serg [3582]

power = 205.8 \: watt \\ solution \\ mass = 66 \: kg \\ time = 44 \: sec \\ distance = 14 \\ now \\ power = \frac{w}{t} \\ \: \: \: \: \: \: \: \: = \frac{f \times d}{t} \\ \: \: \: \: \: = \frac{m \times g \times d}{t} \\ \: \: \: \: \: \: = \frac{66 \times 9.8 \times 14}{44} \\ \: \: \: \: = \frac{9055.2}{44} \\ \: \: \: \: \: = 205.8 \: watt \\ hope \: it \: helps

4 0
1 month ago
An electric toaster is rated 1200 watts at 120 volts. what is the total electrical energy used to operate the toaster for 30 sec
Yuliya22 [3333]
Energy can be determined using the formula power multiplied by time, thus for a power of 1200 W (or 1200 Joules per second) and a duration of 30 seconds, the calculation yields 36000 J or 36 kJ of electrical energy.

If you need the electrical charge or current: Power equals voltage times current. Therefore, with a power of 1200 watts and a voltage of 120 V, the current can be found as 1200 W divided by 120 V, which results in 10 Amperes. The charge is obtained by multiplying 10 A by 30 s, yielding 300 C.
3 0
16 days ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
Maru [3345]

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

7 0
1 month ago
One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the
serg [3582]
<span>Conclusion: The door's weight results in a CCW torque which can be calculated as Tccw = 145 N*3.13 m / 2. You require a CW torque that balances this Tcw = F*2.5 m*sin20</span>
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1 month ago
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