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Misha Larkins [42]
3 years ago
13

Is mass conserved in 2Na+2H2O=2NaOH+H2​

Chemistry
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

yes mass conservation is in 2Na+2H2O=2NaOH+H2​

Explanation:

it tells how many atoms or molecules of a reactant or a product take part in the reaction

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The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e
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Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           N_{2}O_{4} \rightarrow 2NO_{2}

Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

Initial :               x                   0

Change :       - 0.1        2 \times 0.1

Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

     K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

                 2 = \frac{(0.2)^{2}}{(x - 0.1)}

                2 \times (x - 0.1) = (0.2)^{2}

                            x = 0.12

This means that P_{N_{2}O_{4}} = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

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At a certain temperature the vapor pressure of pure heptane (C_7 H_16) is measured to be 170. torr. Suppose a solution is prepar
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Explanation:

The given data is as follows.

       Pressure (P) = 170 torr,      mass of heptane (m) = 86.7 g

First, we will calculate the number of moles as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                        = \frac{86.7 g}{100 g/mol}

                        = 0.867 mol

Now, the number of moles of CH_{3}COBr are calculated as follows.

     No. of moles =  \frac{mass}{\text{molar mass}}

                           =  \frac{125}{122.9}    

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Therefore, mole fraction of heptane will be calculated as follows.

     Mole fraction = \frac{\text{moles of heptane}}{\text{total moles}}

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                           = \frac{0.867}{1.937}

                           = 0.445

Now, we will calculate the partial pressure of heptane as follows.

            P_{A} = x_{A}P^{o}_{A}

                 = 170 \times 0.445      

                 = 75.65 torr

Thus, we can conclude that the partial pressure of heptane vapor above this solution is 75.65 torr.

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