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Firdavs [7]
3 years ago
8

If you breathe out 0.3 moles of carbon dioxide, how many molecules of carbon dioxide did you breathe out?

Chemistry
1 answer:
pochemuha3 years ago
4 0

Explanation:

1 mole of C molecules contain=6.023×10^23

0.3 moles of C=6.023×10^23/1×0.3

=1.8×10^23 of C molecules

sorry..not sure

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you wish to prepare an hc2h3o2 buffer with a ph of 5.44. if the pka of the acid is 4.74, what ratio of c2h3o2-/hc2h3o2 must you
vazorg [7]

The ratio of buffer C₂H₃O₂ /HC₂H₃O₂  must you use are1:0.199 or 10:2

the ratio of buffer C₂H₃O₂ /HC₂H₃O₂  can be calculate using the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:

pH = pKa + log ([base]/[acid]

Where,

[base] = concentration of C₂H₃O₂in molarity or moles

[acid] = concentration of HC₂H₃O₂ in molarity or moles

For the sake of easy calculation, allow us to assume that:

[base] =1

[acid] = x

Therefore using equation 1,

5.44 = 4.74 + log (1 / x)

log [base / acid] = 0.7

1 / x = 5.0118

x = 0.199

The required ratio of buffer C₂H₃O₂ /HC₂H₃O₂ is 1:0.199 or 10:2

learn more about buffer ratio here brainly.com/question/4342532

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5 0
1 year ago
For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in
Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>

<em>C(s) + 2H2(g) <--> CH4(g). </em>

<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

<u><em>PpCH4 = 1.44 atm</em></u>

6 0
4 years ago
HELP ASAP no longer than 30 seconds
stich3 [128]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
Complete the Lewis structures for pyridine (C5H5N) showing the two most common resonance forms. Draw the Lewis structures with t
agasfer [191]

Answer:

Look at the picture.

Explanation:

So on picture two resonance structures are shown. You can see that nitrogen's electron pair is not used in aromatic system and is planar to the ring.

8 0
3 years ago
Read 2 more answers
The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 2.8\times 10^{-3} min⁻¹

Initial concentration [A_0] = 125 kPa

Final concentration [A_t] = 91 kPa

Time = ?

Applying in the above equation, we get that:-

91=125e^{-2.8\times 10^{-3}\times t}

125e^{-2.8\times \:10^{-3}t}=91

-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)

t=113\ min

3 0
3 years ago
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