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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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Nitric oxide, an important pollutant in air, is formed from the elements nitrogen and oxygen at high temperatures, such as those
Tanya [424]

Answer:

Q is 0.5 and K is 0.01, therefore K is less than Q.

the system will proceed in the reverse direction, thereby converting products into reactants. converting Nitric oxide to form oxygen and nitrogen.

Explanation:

Given,

N2 (g) + O2 (g) « 2 NO (g)

K = 0.01

heated temperature = 2000 ⁰C

N2 moles =0.4

O2 moles =0.1

2NO moles = 0.08

Volume of container = 1L

Q is the reaction quotient  of the equilibrium equation. it is use to determined which direction the system will move to by comparing it with the K value.

Q can be calculated by multiplication of mass of reactant divide by mass of product

Q =\frac{ [N2] *[O2}{2NO}

the mass of the chemicals are in mole, there is need to convert them to moles/litre

therefore,

for N₂ : 0.4 mole per 1L container = 0.4 mol/L = 0.4M

for O₂ : 0.1 mole per 1L container = 0.1 mol/L = 0.1M

for 2NO : 0.08mole per 1L container = 0.08 mol/L = 0.08M

Q =\frac{ [N2] *[O2}{2NO} = (0.4*0.1)/0.08 = 0.5

finally, we are going to compare the value of Q with K

Note that:

if K>Q , the reaction will proceed forward, converting reactants into product.

if K<Q , the reaction will proceed backward, converting products into reactants.

if K=Q , the system of reaction is already in equilibrium.

since, Q is 0.5 and K is 0.01, therefore K is less than Q.

the system will proceed in the reverse direction, thereby converting products into reactants. converting Nitric oxide to form oxygen and nitrogen.

7 0
3 years ago
Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the
Alecsey [184]

Explanation:

The chemical equation is as follows.

      \frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)

And, the given enthalpy is as follows.

    \frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g);  \Delta H = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   \Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}

    102.5 = [(\frac{1}{2})x + 498] - [(2)(243)]

    102.5 = (\frac{1}{2})x + 498 - 486

     102.5 - 12 = \frac{x}{2}

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       \Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})

              x = [(\frac{1}{2})181 + (\frac{1}{2})498] - 243

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0
3 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0
3 years ago
The pressure of a gas increases when the temperature _ at constant volume
weqwewe [10]
When it’s warmer so when temperature encreases
7 0
3 years ago
What are families 3-12 on the periodic table called
fredd [130]

Answer:

transition elements

Explanation:

6 0
3 years ago
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