A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O.

Answers

Answer 1
Answer:

The given question is incomplete. But the complete question is this:

A mixture of CH_(4) and H_2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of H_2, 43.0 g of CH_(4), and 48.4 g of H_(2)O. Assuming that equilibrium has been reached, calculate K_(p) for the reaction.

Explanation:

As the given reaction is as follows.

       CH_4 + H_2O \rightarrow CO + 3H_2

And, we know that

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

Therefore, calculate the moles as follows.

      Moles of CH_4 = (43)/(16.04)

                                  = 2.6808  mol

      Moles of H_2O = (48.4)/(18.01528)

                                  = 2.6866 mol

     Moles of CO = (8.62)/(28.01)

                           = 0.307747  mol

       Moles of H_(2) = (2.6)/(2.01588)

                           = 1.2897  mol

As, we know that

        Concentration = (moles)/(volume (L))

Given volume = 5 L

Hence, calculate the concentration of given species as follows.

    Conc. of CH_4 = (2.6875)/(5)

                                 = 0.5361

     Conc. of H_2O = (2.6889)/(5)

                                = 0.5373

      Conc. of CO = (0.307747)/(5)

                                  = 0.06155

and,  Conc. of H_2 = (1.2897)/(5)

                               = 0.2579

Now, expression for equilibrium constant for the given reaction is as follows.

        K_(c) = ([CO][H_2]^(3))/([CH_4][H_2O])

Now, putting the given values into the above formula as follows.

        K_(c) = ([0.06155][0.2579]^(3))/([0.5361][0.5373])

         K_(c) = 3.665 * 10^(-3)

Also, we know that

             K_p = K_c * (RT)^dn

Consider the equation

        CH_4(g) + H2O(g) \rightarrow CO(g) + 3H_2(g)

Calculate change in moles of gas as follows.

         change in gas moles (dn) = 1 + 3 - 1 - 1

                                              dn = 2

As,    K_p = K_c * (RT)^2

It is given that,  

T = 1000 K,         R = 0.0821

So,

         K_p = 3.665 * 10^(-3) * (0.0821 * 1000)^(2)

          K_p = 24.70

Thus, we can conclude that value of K_(p) for the reaction is 24.70.


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What is the pressure in a gas container that is connected to an open-end u-tube manometer if the pressure of the atmosphere is 742 torr and the level of mercury in the arm connected to the container is 8.60 cm higher than the level of mercury open to the atmosphere?

Answers

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Explanation:

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Remember it!Determine the statement that accurately explains the
endothermic and exothermic reactions.
4
O Endothermic reactions absorb heat
energy while exothermic reactions
release heat energy.
O Exothermic reactions absorb heat
energy while endothermic reactions
release heat energy.
O Endothermic and exothermic reactions
absorb thermal energy.
O Endothermic and exothermic reactions
release thermal energy.
REM
The prefix exo-
applied to a che
mean energy
re

Answers

Statement-1 is the correct answer that is Endothermic reactions absorb heat energy while exothermic reactions release heat energy.

How can we distinguish between exothermic and endothermic processes?

Chemical reactions that release energy are referred to as exothermic. More energy is produced than is needed to break the bonds between the reactants when bonds are formed in the byproducts of exothermic processes.

Chemical processes that either use or absorb energy are referred to as endothermic. In endothermic reactions, more energy is absorbed when bonds in the reactants are broken than is released when new bonds are formed in the products. Since an isothermic chemical reaction uses exactly as much energy as it produces, there is no net energy change.

Therefore, the first statement is the right response.

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Answers

You did not specify the types, but I believe the answer would be gamma radiation.

Final answer:

Among alpha,beta, neutron, and gamma radiations, gamma radiation has the greatest ability to penetrate matter. It can pass through dense and thick materials, including a few feet of concrete and even lead.

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When comparing the ability of different types of radiation to penetrate matter, from least penetrating to most penetrating, the sequence is alpha < beta < neutron < gamma. This means the gamma radiation is the most penetrating type of radiation. Alpha radiation has the least penetration ability and can be blocked by even a thin sheet of paper. Beta radiation has slightly higher penetration power, able to go through about 3mm of aluminum, for instance. Gamma radiation, on the other hand, can penetrate much denser and thicker materials. For example, high-energy gamma radiation can pass through a few feet of concrete, and even lead to a depth of 2 or more centimeters.

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