Marine organisms that are well adapted to drastic changes in salinity would most likely be found in which environment?

What volume of air is needed to burn completely 100cm of propane?

Evgesh-ka [11]

Answer:

See below

Explanation:

propane mole weight = 44 gm / mole

100 gm / 44 gm / mole = 2.27 moles

From the equation, 5 times as many moles of OXYGEN (O2)are required

= 11.36 moles of oxygen

at STP this is 254.55 liters of O2 (because 22.4 L = one mole) and

Using oxygen as 21 percent of air means that

.21 x = 254.55 = x = 1212.12 liters of air required

5 0

1 year ago

2 HCl + Na2SO4 > 2 NaCl + H2SO4

Cerrena [4.2K]

Answer:

15

Explanation:

8 0

1 year ago

Which groups on the periodic table contain metalloids? Groups 1–2 Groups 3–12 Groups 13–16 Groups 17–18

agasfer [191]

The correct answer is groups 13-16

6 0

2 years ago

Read 2 more answers

A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a

Andrew [12]

Answer:

Question 1: 1 s after the motion starts

Question 2: 0 (just when the motion starts)

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

1. From the speedometer shown at the right.

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

Free fall equation: Vf = Vo - gt

Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

2. Time when the speedometer displays a reading of 20 m/s

First, calculate the launching speed:

Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

0 = Vo - gt

Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

20 m/s = 30 m/s - 10 m/s² × t

t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

3. Time when the speedometer displays a reading of 30 m/s

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0

2 years ago

After he conducted cathode ray tube experiments proving the existence of negatively charged particles we now call electrons, Tho

Lina20 [59]

Answer:

Answer is explained below;

Explanation:

In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.

In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.

By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.

He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.

So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.

Later, the Rutherford model was replaced by the Bohr atomic model.

6 0

2 years ago