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Amanda [17]
2 years ago
10

Sodium an alkali metal and chlorine a halogen are both in period 3?of the periodic table. Which element has a higher ionization

energy? Explain
Chemistry
1 answer:
V125BC [204]2 years ago
7 0

Answer:

chlorine

Explanation:

Because in periodic table across the period from left to right, ionization energy increases due to increase in the nuclear charge as more and more electrons are added to the same shell.

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What were schools like the first half of Henry's life? How do you think this affect his education? Answer BOTH questions in comp
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Answer:

Not sure.

Explanation:

Who is Henry?

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2 years ago
What is Matter and give the examples​
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Matter is a substance that has inertia and occupies physical space.

Matter is literately in <u>everything</u>

Explanation:

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1 year ago
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1. If you have 25% radioactive parent atoms in a mineral crystal, and 75% daughter atoms, how many half-lives have passed?
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Too freaking many... or maybe not many at all

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3 years ago
Which organism is able to cause an infection
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3 years ago
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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI
NeTakaya

Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 10^{-17}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

CuI\rightleftharpoons Cu^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][I^-]

1\times 10^{-12}=0.0079\times [I^-]

[I^-]=1.25\times 10^{-10}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgI\rightleftharpoons Ag^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][I^-]

8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

8 0
3 years ago
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