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Ann [662]
2 years ago
7

How many molecules of co2 are in a 500. 0 ml container at 780 mm hg and 135°c? 8. 76 × 1021 molecules 9. 23 × 1021 molecules 5.

50 × 1021 molecules 2. 65 × 1022 molecules 2. 79 × 1022 molecules.
Chemistry
1 answer:
Aloiza [94]2 years ago
5 0

<u>Step 1:</u>

ok we have to use the formula PV=nRT

p=Pressure (must be converted to atm)= 780 mmHg

1 amt= 760 mmHg use this as a conversion factor

780 mmHg (1 atm/760 mmHg)= 1.026

V= Volume= 5.00 mL = o.5 L

n=number of moles which we have to find first

R= 0.0821

T(convert to Kelvins by adding 273.15 to the celsius temperature)= 135 C + 273.15= 408.15 k

Now plug in->

(1.026 atm)(o.5 L)= n(0.0821)(408.15 K)

(1.026 atm)(0.5 L)= n(33.509115)

(0.513)= n(33.509115)

n(number of moles)= 0.01532 mol

Now we have to convert to moles using Avagodro's number which states that 1 mol = 6.022 x 10^23 molecules or atoms

So 0.01532 mol (6.022 x 10^23 number of molesules)/ (1 mol) = 9.225704 x 10^21 = 9.226 x 10^21 colecules

Step 2

You must transfer pressure into pascals, 780 mm Hg = 103991 Pa

135*C = 408.15 k

then from the equation pV = nRt

n = pV / RT (T in Kelvins, V in M^3)

n = 103991 x 500 x 10^-6 / (8.314 x 408.15)= 0.015322 moles of N2

1 mol of everything is 6.022 x 10^23 particles, so 0.15322 moles is 0.15322 x 6.022 x 10^23 = 9.2269084 x 10^21 molecules

Explanation:

Hope this helps :)

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Explanation:

Given that:

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 The I.C.E table is as follows:

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I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

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= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

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x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M

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