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The frequency is 91.74 MHz.
It is UltraViolet Radiation.
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2K + Br2 ===> 2KBr
It's very ionic. The transfer of 2 electrons from K to Br2 is nearly as complete as it can be.
The major product of reaction between 1-butanol and Na2Cr2O7 is butanoic acid.
When a primary alcohol like 1-butanol (OH is bonded to a primary carbon) is begin to oxidize in the presence of strong oxidizing reagent such as sodium dichromate (Na2Cr2O7) and H2SO4, sulfuric acid, the stepwise oxidation take place as above firstly to the corresponding aldehyde which undergoes further oxidation to the corresponding carboxylic acid.
You can find that the formed aldehyde after first oxidation is butanal and the only organic product, due to the strong oxidizing reagent is butanoic acid.
Thus, the major product formed is butanoic acid.
learn more about oxidation:
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Answer:
The hydrogen produces the smaller amount of ammonia.
Step-by-step explanation:
We are given the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 28.02 2.016 17.03
N₂ + 3H₂ ⟶ 2NH₃
Mass/g: 70.0 7.00
1. Calculate the moles of N₂ and H₂
Moles N₂ = 70.0 × 1/28.02
Moles N₂ = 2.498 mol N₂
Moles H₂ = 7.00 × 2.016
Moles H₂ = 3.472 mol N₂
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2. Calculate the moles of NH₃ from each reactant
<em>From</em> N₂:
The molar ratio is 2 mol NH₃/1 mol N₂
Moles of NH₃ = 2.498 × 2/1
Moles of NH₃ = 4.996 mol NH₃
<em>From</em> H₂:
The molar ratio is 2 mol NH₃/3 mol H₂
Moles of NH₃ = 3.472 × 2/3
Moles of NH₃ = 4.139 mol NH₃
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3. Identify the limiting reactant
The limiting reactant is H₂, because it produces fewer moles of NH₃.