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masya89 [10]
4 years ago
9

Which has greater mass, 1 mole of iron atoms or 1 mole of lithium

Chemistry
1 answer:
Shkiper50 [21]4 years ago
8 0
1 mole of iron atoms.
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PLEASEEEEEEEEEEEE HELP 100 POINTS PLEASEEEE Sofia has two 10-gram samples of sea salt. One sample is finely ground into thousand
s344n2d4d5 [400]

Answer:

I think it is D.

Explanation:

By the way, there is only 5 points on this question

4 0
3 years ago
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Can someone please help by filling out this table?
tamaranim1 [39]

Answer:

Lithium looses one electron.

The charge - Li^{+} - Cation

Nitrogen gains three electrons - Ni^{3-} - Anion

Boron losses three electron.

The charge - B^{3+} - Cation

Explanation:

Lithium looses one electron.

The charge - Li^{+} - Cation

Nitrogen gains three electrons - Ni^{3-} - Anion

Boron losses three electron.

The charge - B^{3+} - Cation

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a chemist want to make a solution of 3.4M HCl there are tow solutions of HCl that he can find on the shelf. one has a concentrat
steposvetlana [31]

Answer:

The most concentrated one, 6.0 M.

Explanation:

A simple and reliable way to produce a solution of HCl (or anything else, for that matter) is to use a more concentrated solution and dilute it.

In this case the chemist could take a portion of the 6.0 M solution and dilute it by adding water, until the concentration is 3.4 M.

Such a process would not be possible with the 2.0 N (which is the same as 2.0 M for HCl) solution.

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3 years ago
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Rom4ik [11]

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3 years ago
Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What i
AlekseyPX

Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

      AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I                      0              0

C                    +S            +2S

E                      S              2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

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