Moles of H₂ are needed to produce 9.33 moles of NH₃ : 13.995
<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.
The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products
Reaction for the synthesis of ammonia :
N₂+3H₂⇒2NH₃
moles of NH₃ = 9.33
From equation, mol ratio of H₂ : NH₃ = 3 : 2, so mol H₂ :
Answer:
4.14 x 10²⁴ molecules CO₂
Explanation:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O
To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.
4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀
100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀
6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.
Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
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