Electron
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Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
I assume what they are asking you? Sorry if that sound mean
Answer:
The equation is Fe₂O₃ + CO ⇒ Fe + CO₂.
The balanced reaction equation is Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂.
Explanation:
First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.
Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.
First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:
Fe₂O₃ + CO ⇒ 2Fe + CO₂
Next, we need to make sure that we have the same number of C and O atoms on each side. This takes a little bit of thinking, but what we have to do is give CO a coefficient of 3 and CO₂ a coefficient of 3. This gives us 6 O atoms on the left side (when we include the O₃) and 6 O atoms on the right side (since there are 3 O₂ atoms and 3 times 2 is 6). Here's what that looks like:
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
And that's how I balanced the equation. It can be confusing, but with enough practice, it will get easier and easier. :)
The Benzenesulfonic acid does not undergo Friedel-Crafts alkylation because of the deactivation of the compound by the carboxylic group.
<h3>What is the Grignard reagent?</h3>
The Grignard reagent is a compound that contains alkyl magnesium halide.
a) The student will be unsuccessful to prepare a Grignard reagent from 4-bromocyclohexanol because of the -OH group that reacts with the Grignard reagent when formed.
b) The Benzenesulfonic acid does not undergo Friedel-Crafts alkylation because of the deactivation of the compound by the carboxylic group.
c) The compound (2S, 3R)- 2,3-Dibromobutane has a specific rotation, [a]D, 0⁰ because it is a meso compound.
d) This is because, the tertiary alkyl halide is more prone to elimination reaction giving the alkene.
e) This is because, the reaction may be occurring by an SN1 mechanism and the rate determining step is the formation of the carbocation.
Learn more about substitution reaction:brainly.com/question/16811879
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