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ivanzaharov [21]
3 years ago
8

2NaClO3 ——> 2NaCl + 3O2

Chemistry
1 answer:
Alborosie3 years ago
3 0

Answer:  

<h3>1) 208.09 g</h3><h3>2) 7.11 × 10²² particles</h3>

<h2>Explanation:</h2><h3>NUMBER ONE</h3>

<u>Find moles of Known Substance (NaCl)</u>

Moles of NaCl = mass ÷ molar mass

                        =  114.25 g ÷ 58.44 g/mol

                        = 1.955 mol

<u>Use Mole Ratio of Known to Unknown (NaCl : NaClO₃) to Determine Moles of Unknown</u>

The mole ratio of NaCl to NaClO₃ is 2 to 2 or simply 1 to 1.

This means that for each mole of NaCl produced the same number of moles of NaClO₃ was decomposed.

∴ since moles of NaCl = 1.955 mol,

then moles of NaClO₃ = 1.955 mol

<u>Find Mass of Unknown </u>

Mass of NaClO₃ = moles × molar mass

                           = 1.955 mol × 106.44 g/mol

                           =  208.09 g

<h3>NUMBER TWO</h3>

<u>Determine the Moles of Known (O₂)</u>

moles of O₂ = (volume × density) ÷ molar mass

                    = (68.30 L × 0.0828 g/L) ÷ 32 g/mol

                    = 0.177 mol

<u>Use Mole Ratio of Known to Unknown (O₂ : NaClO₃) to Determine Moles of Unknown</u>

Mole ratio of O₂ : NaClO₃ is 3 : 2. Therefore for each mole of oxygen produced, ²/₃ the moles of NaClO₃ is produced.

∴ if moles of O₂  =  0.177 mol

then moles of NaClO₃ =  0.177 mol × ²/₃

                                     = 0.118 mol

<u>Find the Number of Particles of Unknown</u>

Number of particles  = moles × Avogadro's Number

                                  = 0.118 mol × (6.022 × 10²³ particles/mol)

                                  =  7.11 × 10²² particles

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stiks02 [169]

Answer:

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Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

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7 0
4 years ago
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3 years ago
a 5.00 l air sample has a pressure of 107 kpa at a temperature of -50.0 C. if the temperature is raised to 102 C and the volume
Murljashka [212]
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4 0
3 years ago
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Vesna [10]

Answer:

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Explanation:

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Half-reaction: O_{2}(g)\rightarrow H_{2}O_{2}(aq.)

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7 0
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vesna_86 [32]

Answer:

False

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Hence , the reaction involving acid anhydride  are conducted in anhydrous solvents .

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5 0
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