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Yuri [45]
3 years ago
8

Consider the following balanced equation:

Chemistry
1 answer:
algol [13]3 years ago
8 0

Moles of PF₃ : 4

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

\tt P_4(s)+6F_2(g)\rightarrow 4PF_3(g)

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)

Limiting reactant : the smallest ratio (mol divide by coefficient)

P₄ : F₂ =

\tt \dfrac{1.25}{1}\div \dfrac{6}{6}=1.25\div 1\rightarrow F_2~limiting~reactant(smallest~ratio)

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

\tt \dfrac{4}{6}\times 6~moles=4~moles

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a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas
jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

Partial pressure of SO2 = 0.409 atm

Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

pO2 = 0.601 atm

pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

 X = 0.056 atm

pO2 = 0.601 - 0.056 = 0.545 atm

pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

3 0
3 years ago
What mass of compound must be weighed out to have a 0.0223 mol sample of H2C2O4 (M=90.04 g/mol)?
rodikova [14]
Mass of  H2C2O4 :

mm = 90.04 g/mol

number of moles : 0.0223 moles

m = n * mm

m = 0.0223 * 90.04

m = 2.007 g 

hope this helps!.

5 0
3 years ago
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Answer:

7.0

Explanation:

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How many moles are in 337 grams of tellurium?
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MA_775_DIABLO [31]
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Now C₍₅ₓ₂₎H₍₁₁ₓ₂) = C₁₀H₂₂

Hope that helps
6 0
3 years ago
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