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Natalija [7]
3 years ago
8

Write the equilibrium expression for N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) + heat I have no idea where to even start, due to the corona

we have to learn by ourselves and my teacher is slacking off. Please Help!
Chemistry
1 answer:
motikmotik3 years ago
5 0

Answer:

The expression of an equilibrium constant will given as:

K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium

N_2 (g) + 3 H_2 (g)\rightleftharpoons 2 NH_3 (g) + heat

The expression of an equilibrium constant will given as:

K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}

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Vikentia [17]

Upwelling

Explanation:

The rise of deeper water to the surface is called upwelling.

Upwelling is a term used in oceanography to describe the rise of cold and denser water to the surface.

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3 0
3 years ago
How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has
Papessa [141]

The  balanced chemical reaction is given as:

2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)

Now, convert 19.0 ft^{3} into litres.

1 ft^{3}  = 28.3168

So, 19.0 ft^{3} = 19\times 28.3168 = 538.0192 L

Density is equal to the ratio of mass to the volume.

D=\frac{M}{V}

where, M = mass and V= volume (538.0192 L)

Substitute the value of density and volume in formula to get the value of mass.

1.25 g/L=\frac{M}{538.0192 L}

1.25 g/L\times 538.0192 L= M

Mass = 672.524 g

Now, number of moles of N_{2} gas=\frac{672.524 g}{28.02 g/mol}

= 24.00 moles

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= \frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles of nitrogen gas, moles of sodium azide.

number of moles of sodium azide  = 16 moles

Mass of sodium azide in g  =  number of moles\times molar mass of sodium azide.

= 16 moles\times 65.00 g/mol

= 1040 g

Thus, mass of sodium azide which is required to produce 19.0 ft^{3} of nitrogen gas  = 1040 g





3 0
3 years ago
A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

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3 years ago
What is ancient japan remembered for
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Hi!

The Samurai is what I would attribute to the memory of Ancient<span> Japan.</span>
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Match the following descriptions with the correct polysaccharide, note if you answer any part of this question incorrectly a sig
Ad libitum [116K]

Answer:

Glycogen. Cellulose. Amylose. Cellulose. Amylopetin and Glycogen. Amylopetin and Cellulose.

Explanation:

Glycogen is the form that glucose is stored in human body.

Cellulose is the structural part of plant cell walls and human cannot digest it.

Amylose is the polysaccharide linked mainly by the the bonds of \alpha 1,4 glycosidic.

Cellulose is an unbranched polysaccharide linked mainly by the bonds of  \beta 1,4 glycosidic.

Amylopetin and Glycogen are branched polysaccharides linked by the bonds of \alpha 1,4 glycosidic and \alpha 1,6 glycosidic.

Amylopetin and Cellulose are mainly stored in plants.

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4 years ago
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