What is the mass of a 1.75l sample of a liquid that has a density of 0.921 g/ml?

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

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.

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

O sea,

Dado que:

Para :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de son:

Moles de = 0.004 moles

Para :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de son:

Moles de = 0.012 moles

Según la reacción:

1 mol de reacciona con 1 mol de

Por lo tanto,

0.012 mol de reacciona con 0.012 mol de

Moles disponibles de = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de reacciona con 1 mol de y produce 1 mol de

0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de

Los moles restantes de son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, , después de la reacción es:

Yasir wished to explore how sleep relates to room color. He conducted necessary preliminary research and formed a hypothesis suggesting individuals doze off faster in blue-painted rooms compared to those painted in yellow. Yasir surveyed several individuals about their color preference—yellow or blue—and utilized their feedback to assess the validity of his hypothesis. However, he did not conduct an actual experiment to examine the impact of room color on sleep, and he failed to clearly define the variables that should have been part of his experiment.

Hence, the correct answer would be,

An experiment that directly tests the hypothesis

Variables to be tested by an experiment