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kvv77 [185]
2 years ago
12

Use the periodic table to identify the number of core electrons and the number of valence electrons in

Chemistry
1 answer:
deff fn [24]2 years ago
6 0

Answer:

10 core electrons and 8 balance electrons

Explanation:

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Give the answer to the problem below using the correct number of significant digits. (1.3 x 103) x (5.724 x 104)
Gemiola [76]
(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.
6 0
3 years ago
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Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
Explain why the electron configuration of 2-3-1 represents an atom in an excited state.
Allushta [10]

Answer:

The excited state electron configuration of an atom indicates the promotion of a valence electron to a higher energy state.

4 0
3 years ago
The amount of energy available to do work after a chemical reaction has occurred is called
sveticcg [70]
Free energy is the answer i hope this helped
8 0
3 years ago
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
3 years ago
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