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nlexa [21]
3 years ago
14

What column(s) have 5 valence electrons?

Chemistry
1 answer:
kodGreya [7K]3 years ago
6 0

Answer:

Group 15

Explanation:

The elements of group 15 (column) VA of the periodic table all have electron configurations of s2p3 , giving them five valence electrons. These elements include Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb) and Bismuth (Bi).

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For each solution, determine the p-values for each ion indicated. A solution that is 0.493 M in NaCl and 0.314 M in NH 4 Cl .
miss Akunina [59]

Complete Question:

Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺

Answer:

pNa = 0.307

pCl = 0.093

pNH₄ = 0.503

Explanation:

The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.

Both substances are salts that solubilize completely, thus, by the solution reactions:

NaCl → Na⁺ + Cl⁻

NH₄Cl → NH₄⁺ + Cl⁻

So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.

[Na⁺] = 0.493 M

[Cl⁻] = 0.493 + 0.314 = 0.807 M

[NH₄⁺] = 0.314 M

The p-values are:

pNa = -log[Na⁺] = -log(0.493) = 0.307

pCl = -log[Cl⁻] = -log(0.807) = 0.093

pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503

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3 years ago
What is a device that changes electrical energy into mechanical energy?
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A conversion factor set up correctly to convert 15 inches to cm is _______.
sleet_krkn [62]

Answer:

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6 0
3 years ago
Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how
Afina-wow [57]

Answer:

11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

Explanation:

1. The balanced chemical equation is the following:

Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)

2. Use the molar mass of the H_{2}O, the molar mass of the NH_{3} and the stoichiometry of the balanced chemical reaction to find how many grams of NH_{3} are produced:

Molar mass H_{2}O = 18\frac{g}{mol}

Molar mass NH_{3} = 17\frac{g}{mol}

36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}

Therefore 11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

3 0
3 years ago
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