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hram777 [196]
3 years ago
11

What is the Kp of this reaction at 700oC if the Kc is 0.40? 

Chemistry
2 answers:
m_a_m_a [10]3 years ago
8 0

Answer:

The answer is B) 6.27 x 10⁻⁵

Explanation:

The relation between Kp and Kc is given by the following equation:

Kp= Kc (RT)ⁿ

We have:

Kc= 0.40

T= 700ºC= 973K

R=0,082 L.atm/K.mol

To obtain n (or usually called Δn which is the change in mol), we have to look at the number of mol of products (CH₃OH) and reactants (CO and H₂) in gaseous phase.

n= Δn= (total number of mol of products) - (total number of mol of reactants)

n=Δn= (mol CH₃OH) - (mol CO + mol H₂)= 1 - (1+2)= 1-3= -2

So, we introduce the values in the equation to obtain Kp:

Kp= Kc (RT)ⁿ

Kp= 0.40 (0.082 x 973)⁻²

Kp= 6.27 x 10⁻⁵

dem82 [27]3 years ago
3 0
I am pretty sure its D
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Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

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The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

4 0
3 years ago
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evablogger [386]

Answer: Option (c) is the correct answer.

Explanation:

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Thus, we can conclude that chemical energy is stored in our food and this chemical energy start out as light energy from the sun.

6 0
3 years ago
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A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

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3 years ago
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3 years ago
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[NO₂]²/[N₂O₄]

Hope that helps.

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