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KengaRu [80]
3 years ago
5

onsider the reaction, which takes place at a certain elevated temperature CO(g)+NH3(g)⇌HCONH2(g), Kc=0.860 If a reaction vessel

initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium?
Chemistry
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

<h2> concentration of HCONH2 at equillibrium is 0.5 M</h2>

Explanation:

                               CO(g) + NH3(g) ⇌ HCONH2(g)

-vessel iniatial             1             2                 0

concentration

-equillibrium              1-x         2-x               x

concentration

                                K_{C} = (CO)\times (NH_{3})/ (HCONH_{2})

                               0.860 = (1-x)(2-x) / x

                                 x^{2}-3.86x+2

                                  x = 3 , 0.5

         concentration of HCONH2 at equillibrium is 0.5 M

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Answer:

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Explanation:

3 0
3 years ago
What is the final concentration of the solution produced when 225.5 mL of 0.09988-M solution of Na2CO3 is allowed to evaporate u
ikadub [295]

Explanation:

The given data is as follows.

       V_{1} = 225.5 mL,             V_{2} = 45.00 mL

       M_{1} = 0.09988 M,          M_{2} = ?

Therefore, formula to calculate final concentration will be as follows.

                  M_{1}V_{1} = M_{2}V_{2}

Putting the given values into the above formula as follows.

            M_{1}V_{1} = M_{2}V_{2}

   0.09988 M \times 225.5 mL = M_{2} \times 45.0 mL

                    M_{2} = 0.5 M

Thus, we can conclude that the final concentration of the given solution is 0.5 M.

4 0
3 years ago
During one researcher's experiment, the total mass of the gas-generating solid and entire assembly was 52.1487g before the react
irinina [24]

Answer:

43.93 g/mol

Explanation:

The mass of the gas before reaction = 52.1487 g

The mass of the gas after reaction = 52.1098 g

Mass of gas generated = 0.0389 g

Moles of the gas = 8.854\times 10^{-4}\ moles

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Molar\ mass = \frac{Mass\ taken}{Moles}

Molar\ mass= \frac{0.0389\ g}{8.854\times 10^{-4}\ moles}

Molar mass of the gas = 43.93 g/mol

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Hope this helps!
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3 years ago
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