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Nataly_w [17]
2 years ago
13

While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti

es but with a different atomic mass. Which best describes the student’s options? There are two possible elements, and they are directly to the left and right of bromine. There are two possible elements, and they are directly above and below bromine. There are many possible elements, and they are all in the same vertical column as bromine. There are many possible elements, and they are all in the same horizontal row as bromine.

Physics
2 answers:
Ede4ka [16]2 years ago
7 0

Answer: There are many possible elements, and they are all in the same vertical column as bromine.

Explanation:

In a periodic table, the elements are arranged according to the atomic number. The elements arranged in the same vertical column (known as groups) have same valence configuration and therefore have same chemical properties. Hence, there would be more possible elements having same chemical properties in the same vertical column (group) as Bromine.

d1i1m1o1n [39]2 years ago
4 0

Answer:  There are many possible elements, and they are all in the same vertical column as bromine

Explanation: All the elements in the same vertical columns or same groups have similar chemical properties due to the presence of similar valence shell configurations but have different masses as the number of protons and neutrons keep on increasing on moving down the group.

All the halogens (members of fluorine group) are short of one electron each to attain their stable noble gas configuration and hence behave similarly.

[X]:ns^2np^5

The elements present in the same horizontal row or same period differ in the chemical properties as they have different valence shell configurations.

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According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
2 years ago
Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th
prohojiy [21]
The equation to be used is written as:

ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol

r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³
7 0
2 years ago
Read 2 more answers
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
1 year ago
To avoid overdriving your headlights, __________ A. cool your headlights periodically by switching to your parking lights. B. ma
IceJOKER [234]

Answer: The correct answer is A.

Explanation:

To avoid over driving your headlights, cool your headlights periodically by switching to your parking lights.

It is difficult to stop in the distance you see ahead of you at night without using headlights. Otherwise, there may be chances of occurring an accident.

Avoid to use the high beams is used whenever you drive at night. Suppose, if the car is coming in front of your car then in this case the drivers to have a flash blindness. As he gets bright light in his eyes.  Otherwise, there may be chances of occurring an accident.

Therefore, to avoid over driving your headlights, cool your headlights periodically by switching to your parking lights.

6 0
2 years ago
Read 2 more answers
A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this
noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

W=Fdcos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

The displacement of the object is d = 3.00 m, while \theta=180^{\circ}, because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J

And the work done is negative, because it is done against the motion of the object.


6 0
2 years ago
Read 2 more answers
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