A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.

Response:

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

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Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Therefore, the volume of liquid nitrogen is 1.68 L.

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:

vHe / vNe = 2.24

Explanation:

To determine the velocity of an ideal gas, one should apply the formula:

v = √3RT / √M

In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio simplifies to:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it assists you!