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777dan777 [17]
2 years ago
8

If a negatively charged object is brought near a neutrally charged sphere, what will happen? The electrons will move toward the

charged object. The protons will jump to the negatively charged object making both objects neutral. The electrons will jump to the charged object. The electrons will move away from the charged object.
Chemistry
2 answers:
vampirchik [111]2 years ago
7 0

The protons will jump to the negatively charged object making both objects neutral.

Sidana [21]2 years ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

When a charged object is brought near a neutral atom then the neutral atom will develop an opposite charge. This process of inducing charge is known as induction.

Hence, when a negatively charged object is brought near a neutrally charged sphere then a positive charge will develop at the sphere. Thus, there will be attraction because opposite charges attract each other and like charges repel each other.

Hence, we can conclude that if a negatively charged object is brought near a neutrally charged sphere, then the electrons will move toward the charged object.

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5 0
2 years ago
Read 2 more answers
7) Convert 8.34 x 10^-6 Kg to cg"
marusya05 [52]
The answer is .00834 cg hopefully this helped
5 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
This data was collected after conducting an experiment about the amount, in liters, of water a specific plant needs per month. A
Effectus [21]

Answer:

1.1 liters

1.2 liters

1.5 liters

Explanation:

Precision in data refers to how close the experimental values of an experiment are to one another irrespective of the true or accepted value. In other words, a set of values are said to be PRECISE if they are close to one another.

In this case, data was collected after conducting an experiment about the amount, in liters, of water a specific plant needs per month. However, according to the set of experimental values provided, only 1.1 litres, 1.2litres and 1.5litres are close to one another and, hence, are said to be PRECISE even if they are not close to the accepted value of 6litres.

4 0
3 years ago
Read 2 more answers
I need help!! i need to make it like 3:2
anyanavicka [17]

Answer:

use visual studio code and put in this print{3:2}-1

Explanation:

4 0
2 years ago
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