The protocol for a given lab experiment specifies that you should prepare 100 mL of a 0.10 M (mol/L) aqueous solution of NaNO3.

To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C

Answer:

The generation of static electricity occurs when two surfaces are rubbed together. This process causes a transfer of electrons, resulting in a build-up of negative charge. For instance, when you shuffle on a carpet, the friction creates multiple contact points which allow electrons to move onto you, thus accumulating a static charge. Touching another individual or object can lead to a sudden discharge, experienced as an electric shock.

In a similar way, rubbing a balloon against your hair generates opposite static charges on both your hair and the balloon. As you gently pull the balloon away from your head, the attraction between these opposite charges can be observed, causing your hair to rise.

Materials

• Balloon

• Woolen item (like a sweater, scarf, or yarn ball)

• Stopwatch

• Wall

• Partner (optional)

Preparation

• Inflate the balloon and secure the end.

• Have your partner ready to time with the stopwatch.

Procedure

• Grip the balloon with minimal hand coverage, such as holding it with just your thumb and index finger, or by its tied neck.

• Rub the balloon on the wool item once, making sure to go in one direction only.

• Press the rubbed side of the balloon against the wall and let go. Is it adhering to the wall? If it's stuck, your partner should start the stopwatch to measure how long it stays there. If it doesn’t stick, continue to the next step.

• Briefly touch the balloon to a metal object. Why is this step necessary?

• Repeat this procedure, but each time increase the number of rubs against the woolly item, ensuring the direction remains the same (do not rub back and forth).

Observations and results

As you increase the number of times you rub the balloon on the woolly material, does the duration of its adhesion to the wall increase?

Wool is an excellent conductor; it easily relinquishes electrons. When you rub wool on a balloon, electrons move from the wool to the surface of the balloon, imparting a negative charge to the rubbed area. Balloons, made from rubber, act as insulators, which means not all areas of the balloon will have a negative charge—only where it was rubbed will have a negative charge, while the rest of the balloon remains neutral.

Once the balloon is sufficiently charged negatively by repeated rubbing, it will adhere to the wall. Though the wall typically has a neutral charge, its internal charges can realign such that a positively charged region can attract the negatively charged balloon. Since the wall is also an insulator, the charge does not dissipate instantly. However, when the balloon is in contact with a metal object, the excess electrons from the balloon flow into the metal quickly, making the balloon lose its attraction and peel away.

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Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).