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Kaylis [27]
2 years ago
6

At a particular temperature, Kp-483 × 102 for the reaction H2 (g) + 12(g) ? 2H1(g) If 3 01 atm of H-(g) and 3.01 atm of I2(g) ar

e introduced into a 1.00-L- container, calculate the equilibrium partial pressures of all Partial pressure of H2 Partial pressure of I2atm Partial pressure ofHIatm atm
Chemistry
2 answers:
Ira Lisetskai [31]2 years ago
8 0

Answer:

P[H2] = P[I2] = 0.02 atm

P[HI] = 5.98 atm

Explanation:

The ICE table for the given reaction is:

                H2(g)   +      I2(g)      ↔     2HI(g)

Initial       3.01atm       3.01atm            -

Change    -x                 -x                    +2x

Equilib     (3.01-x)        (3.01-x)              2x

The equilibrium constant, Kp is given as:

Kp = \frac{P_{HI}^{2}}{P_{H2}*P_{I2}}

483*10^{2} = \frac{(x)^{2}}{(3.01-x)^{2}}

x = 2.99 atm

Equilibrium partial pressure of H2 = I2 = 3.01-2.99 = 0.02 atm

Equilibrium partial pressure of HI = 2x = 2(2.99) = 5.98 atm

ale4655 [162]2 years ago
8 0

Answer:

The partials pressures are:

PHI = 5.016 atm

PH2 = PI2 = 0.502 atm

Explanation:

please look at the solution in the attached Word file

Download docx
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A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final
salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

M_1V_1=M_2V_2

where,

M_1 = concentration of HNO_3 = 13.5 M

M_2 = concentration of diluted solution = ?

V_1 = volume of HNO_3 = 25.0 ml  = 0.0250 L

conversion used : (1 L = 1000 mL)

V_2 = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

13.5M\times 0.0250L=M_2\times 0.500L

M_2=0.675M

Therefore, the concentration of the diluted solution is 0.675 M

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3 years ago
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3 years ago
Equation is balanced already
mojhsa [17]

Answer:

1.) 13 g C₄H₁₀

2.) 41 g CO₂

Explanation:

To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

 48 g O₂             1 mole             2 moles C₄H₁₀            58.124 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  13 g C₄H₁₀

 48 g O₂             1 mole               8 moles CO₂            44.007 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  41 g CO₂

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