Answer:
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
Explanation:
The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.
Explanation:
Hydrogen (H)
Helium (He)
Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)
Neon (Ne)
Sodium (Na)
Magnesium (Mg)
Aluminum (Al)
Silicon (Si)
Phosphorus (P)
Sulfur (S)
Chlorine (Cl)
Argon (Ar)
Potassium (K)
Calcium (Ca)
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Incorrect, temperature is directly proportional to the avg. KE of a gas.
Answer:
We'll have 82 moles ZnO and 41 moles S
Explanation:
Step 1: data given
Number of moles Zinc (Zn) = 82 moles
Number of moles sulfur oxide (SO2) = 42 moles
Step 2: The balanced equation
2Zn + SO2 → 2ZnO + S
Step 3: Calculate the limiting reactant
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles
There will remain 42-41 = 1 mol SO2
Step 4: Calculate moles of products
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)
For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur
We'll have 82 moles ZnO and 41 moles S
<>"One such trend is closely linked to atomic radii -- ionic radii. Neutral atoms tend to increase in size down a group and decrease across a period. When a neutral atom gains or loses an electron, creating an anion or cation, the atom's radius increases or decreases, respectively."<>
