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Jet001 [13]
3 years ago
6

How many valence electrons are in the outermost shell of all alkali metals?

Chemistry
2 answers:
saul85 [17]3 years ago
7 0

Answer:

A.) 1 is your answer.

Explanation:

The alkali metals, found in group 1 of the periodic table are very reactive metals that do not occur freely in nature. These metals have only one electron in their outer shell. Therefore, they are ready to lose that one electron in ionic bonding with other elements.

A.) 1 is your answer.

viktelen [127]3 years ago
6 0

Answer:

The correct answer is option A.) 1

Explanation:

Hello!

Let's solve this!

Alkali metals are found in group 1 of the periodic table.

The group number indicates the amount of electrons in the last valence layer.

These elements are prone to form ionic junctions.

As they are in group 1, the correct answer is that they have 1 electron in the valence layer.

We conclude that the correct answer is option A.) 1

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The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?
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Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.
6 0
4 years ago
Read 2 more answers
Please help me, I am confused
Solnce55 [7]

Answer:

B

Explanation:

Ionic compound can conduct electricity

4 0
3 years ago
Chuck wants to know how many electrons in an atom are not paired up. Which model would be best for Chuck to
loris [4]

Answer:

D. an orbital notation of the atom

Explanation:

Orbital notiation uses lines and arrows to show shells, subshells, and orbitals for electrons in an atom. Since it shows arrows being paired up in this diagram it would be the best model for Chuck to use.

4 0
3 years ago
The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was
Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}

[Ca^{2+}]=0.007118 \ M

[Mg^2+] = \dfrac{0.003474 \ M\times (26.23  - 20.49 )mL}{10.0 \ mL}

=0.001994 \ M

Mass of Ca²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}

= 0.1598 g

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= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0
3 years ago
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