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3 years ago

How did Democritus contribute to the development of the atom?

1 answer:

6 0

Democritus developed the atomic model. He theorized that atoms were specific to the material which they composed. He also believed that the atoms different in size and shape, were in constant motion in a void, collided with each other; and during these collisions, could stick together.

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Help will give brainlyness

Morgarella [4.7K]

Answer:

The high melting point is also consistent with its description as an ionic solid. In the crystal, each S2− ion is surrounded by an octahedron of six Ca2+ ions, and complementarity, each Ca2+ ion surrounded by six S2− ions.

Explanation:

You can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a 2+ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a 2+ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized Ca2+. The name of a metal ion is the same as the name of the metal atom from which it forms, so Ca2+ is called a calcium ion.

When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1− charge; atoms of group 16 gain two electrons and form ions with a 2− charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1− charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized Br−. (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)

7 0

3 years ago

Assume that a hydrogen atom's electron has been excited to the n ???? 5 level. how many different wavelengths of light can be em

hichkok12 [17]

n = 5 --> n = 4

n = 4 --> n = 3

n = 3 --> n = 2

n = 2 --> n = 1

<span>So we can see that there are 4 different wavelength of light that can be emitted</span>

3 0

4 years ago

Read 2 more answers

Whats the answer and how you got it please :-)

marta [7]

For the element chlorine to be "happy", it needs 8 valence electrons meaning it needs 8 electrons on its outer shell. So the answer would be chlorine because it has 7 valence electrons in the picture and it needs one more to be considered stable.

4 0

4 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of

Serhud [2]

Answer: 0.17 m $CH_3COONH_4$ : Highest freezing point

0.20 m $CoSO_4$ : Second lowest freezing point

0.18 m $MnSO_4$ : Third lowest freezing point

0.42 m ethylene glycol: Lowest freezing point

Explanation:

Depression in freezing point is a colligative property which depend upon the amount of the solute.

$\Delta T_f=i\times k_f\times m$

where,

$\Delta T_f$ = change in freezing point

i= vant hoff factor

$k_f$ = freezing point constant

m = molality

a) 0.17 m $CH_3COONH_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for $CH_3COONH_4$ , thus total concentration will be 0.34 m

b) 0.18 m $MnSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $MnSO_4$ , thus total concentration will be 0.36 m

c) 0.20 m $CoSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $CoSO_4$ , thus total concentration will be 0.40 m

d) 0.42 m ethylene glycol

For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m

The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.

4 0

3 years ago

In a certain electrolysis experiment, 1.24 g of Ag were deposited in one cell (containing an aqueous AgNO3 solution), while 0.65

eduard

Answer:

The correct answer is 169.56 g/mol.

Explanation:

Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,

= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)

= 0.0115 mole of electron

The half cell reaction for the metal X is,

X^3+ (aq) + 3e- = X (s)

From the reaction, it came out that 3 faraday will reduce one mole of X^3+.

The molar mass of X will be,

= 0.650 g/0.0115 *3 mol electron/1 mol

= 56.52 * 3

= 169.56 g/mol

7 0

3 years ago