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VLD [36.1K]
3 years ago
6

Consider the reaction co(g)+nh3(g)âhconh2(g), kc=0.900 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? express your answer with the appropriate units.
Chemistry
2 answers:
Murrr4er [49]3 years ago
6 0
The answer is 254 units 

Cerrena [4.2K]3 years ago
5 0

Answer:

See explanation below

Explanation:

Let's write the equilibrium reaction:

CO(g) + NH3(g) <------> HCONH2(g)   Kc = 0.9

We know the initial concentrations which are 1 and 2 M. In order to know the concentration of equilibrium, we need to do an ICE chart, and then, write the equilibrium expression for this reaction to calculate the concentration.

The ICE chart:

       CO(g) + NH3(g) <------> HCONH2(g)   Kc = 0.9

i)        1              2                         0

c)       -x           -x                          x

e)      1-x          2-x                        x

The equilibrium expression:

Kc = [HCONH2] / [CO][NH3]

Replacing the values we have:

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

0.9(1-x)(2-x) = x

0.9(x²-3x+2) = x

0.9x² - 2.7x + 1.8 = x

0.9x² - 3.7x + 1.8 = 0

From here, we solve for x using the general formula for a quadratic expression:

x = -b ±√b² - 4ac / 2a

Replacing the values here:

x = 3.7 ±√(3.7)² - 4 * 0.9 * 1.8 / 2*0.9

x = 3.7 ±√13.69 - 6.48 / 1.8

x = 3.7 ± 2.69 / 1.8

x1 = 3.7 - 2.69 / 1.8 = 0.56

x2 = 3.7 + 2.69 / 1.8 = 3.55

As 3.55 is > 1 and 2 from the initial concentrations, the correct value is x1 therefore:

[HCONH2] = 0.56 M

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3 years ago
A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give
8_murik_8 [283]

<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}

Where,

m_{solute} = Given mass of solute (KOH) = 3.301 g

M_{solute} = Molar mass of solute (KOH) = 56.1 g/mol

W_{solvent} = Mass of solvent = 96.699 g

Putting values in above equation, we get:

\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m

Hence, the molality of potassium hydroxide solution is 0.608 m

4 0
3 years ago
Calculate the energy of a photon with a frequency of 2.36 x 10-19 Hz.
OlgaM077 [116]

Answer:

Explanation:

energy for photon is calculated in same wasy as for electromagnetic radiation

energy for electromagnetic radiation = hf

where f is the frequency of photon

h is Planck's constant = h = 4.14 × 10−15 eV · s.

thus

energy of photon = 4.14 × 10−15 eV · s *  2.36 x 10-19 Hz

energy of photon = 9.77 * 10−(-15+ -19) eV

energy of photon = 9.77 * 10−34eV        answer

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liubo4ka [24]

Answer:

the volume is 12.79ml

Explanation:

12.8+12.7+12.78+12.88=51.16

51.16/4

=12.79

8 0
3 years ago
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