Question:
Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:
equations
Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.
Answer:
NO
It is present but not consumed
NO Lowers the activation energy of the reaction
Explanation:
A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction
Therefore, whereby NO is not consumed, it is the catalyst
It functions by lowering the activation energy
Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
Which of these is an isoelectronic series? 1) na+, k+, rb+, cs+ 2) k+, ca2+, or, s2– 3) na+, mg2+, s2–, cl– 4) li, be, b, c 5) n
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An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:
Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6
A. Na⁺: 11-1 = 10 electrons
K⁺: 19 - 1 = 18 electrons
Rb⁺: 37-1 = 36 electrons
B. K⁺: 19 - 1 = 18 electrons
Ca²⁺: 20 - 2 = 18 electrons
S²⁻: 16 +2 = 18 electrons
C. Na⁺: 11-1 = 10 electrons
Mg²⁺: 12 - 2 = 10 electrons
S²⁻: 16 +2 = 18 electrons
D. Li=3 electrons
Be=4 electrons
B=5 electrons
C=6 electrons
The answer is letter B.