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SashulF [63]
3 years ago
7

CO(g) + 3 H2(g) → CH4(k) + H2O(g) + heat

Chemistry
1 answer:
weqwewe [10]3 years ago
8 0

Explanation:

AH is negative because the reaction gives out heat (exothermic).

AS is negative because the total entropy of the reactants is greater than that of the products (there are 4 moles of gas on reactant side compared to only 1 mole on the product side). Thus, the reactants are said to be "messier" than the products.

AG is negative at lower temperatures because at low temperatures, -TAS is more negative, resulting in the negative sign of AG.

Hope this helps.

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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T
shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

                C = \frac{P}{K_{h}}

where,   pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at 25^{o}C are as follows.

  p(O_{2}) = 0.21 atm = 0.21 bar

   p(N_{2}) = 0.78 atm = 0.78 bar

   K_{h} for O_{2} = 7.9 \times 10^{2} bar/mol

    K_{h} for N_{2} = 1.6 \times 10^{3} bar /mol

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

      C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L  

      C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L

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

         n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3 mol

         n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3 mol

As the molar mass of O_2 = 32 g/mol

Hence, mass of oxygen will be as follows.

         Mass of O_2 = 32 \times 1.995 \times 10^-3 \times 1000

                           = 63.84 mg

As the molar mass of N_{2} = 28

       Mass of N_{2} = 28 \times 3.66 \times 10^-3 = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg  and nitrogen is 102.5 mg.

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Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

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