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vredina [299]
2 years ago
12

A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

y moles of oxygen have to react with carbon monoxide in order to decrease the overall pressure in the flask by 10.00%? Assume ideal gas behavior. The reaction of carbon monoxide and oxygen gas is described by the following equation. 2CO(g)+O2(g)⟶2CO2(g)
Chemistry
1 answer:
emmasim [6.3K]2 years ago
7 0

Answer:

0,400 moles of Oxygen

Explanation:

Using PV = nRT it is possible to obtain the initial pressure of the flask before the reaction thus:

P = nRT/V

Where:

n are moles (4,000 moles, 2,000 of CO and 2,000 moles of H₂O)

R is gas constant (0,082atmL/molK)

T is temperature (300,0K)

V is volume (0,2000L)

Replacing, <em>P = 492,0 atm</em>

If you want to decrease the pressure in 10,00%, the final pressure must be:

492,0atm - 49,2 atm = 442,8 atm

Solving under the same conditions with this pressure, moles must be:

n = PV/RT

<em>n = 3,600 total moles</em>

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The moles you will have are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

<em>Where X are moles that react</em>

Thus, total moles are:

4,000moles - X = 3,600 moles

X = 0,400 moles

That means that moles of oxygen that have to react are <em>0,400 moles of Oxygen</em>

<em />

I hope it helps!

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Suppose that the microwave radiation has a wavelength of 12.4 cm . How many photons are required to heat 255 mL of coffee from 2
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Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ /  12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

mass of coffee = 255 x 1 = 255 g

heat required to heat up coffee = mass x specific heat x rise in temp

= 255 x 4.18 x ( 62-25 )

= 39438.3 J  .

No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

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5 0
2 years ago
Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because
olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

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