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Zepler [3.9K]
3 years ago
7

Which of the following are always larger than the neutral atoms from which they are formed?

Chemistry
2 answers:
Sedbober [7]3 years ago
7 0

negetive ions took the test

frosja888 [35]3 years ago
4 0
<span>A. positive ions : Positive ions are formed when a neutral atom loses electrons. When neutral atoms lose electrons, the ions formed are always smaller than the neutral atoms.


B. negative ions : Yes. When the neutral atoms gain electrons to form negative ions, they always become larger, because the addition of one electron increases the electrostatic repulsion of the outermost electrons, forcing them to drift apart.


C. cations  No. Cation is the same that positive ion.

D. none of the above </span>
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How many bromine are in 6NaBr?
OleMash [197]
6 sodium and 6 Bromine in 6NaBr
5 0
3 years ago
Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?
german

Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) AlCl_3

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

So, Van't Hoff factor = Number of solute particles = K^{+}+I^{-} = 1 + 1 = 2

(C) CaCl_2

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

So, Van't Hoff factor = Number of solute particles = Ca^{2+}+2Cl^{-} = 1 + 2 = 3

(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

7 0
3 years ago
Use the periodic table to write the electron configuration for rubidium (Rb) in noble has notation
Alexxx [7]

Answer:

Rb: [Kr] 5s  

Step-by-step explanation:

Rb is element 37, the first element in Period 5.

It has one valence electron, so its valence electron configuration is 5s.

The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.

The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.

4 0
3 years ago
Someone help me please.. I will mark as brainliest I promise...​
Kobotan [32]

Answer:

(a) proton

(b) neutron

(c) electron

particles in nucleus are proton and neutron.

atom is electrically neutral because no.of proton= no.of electron=6

5 0
3 years ago
25<br> #of<br> protons<br> # of<br> neutrons<br> # of<br> electrons
Molodets [167]

Answer:

16

Explanation:

number of electrons should also be 16.

5 0
3 years ago
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