Answer:
I believe is the correct answer
Explanation:
100bromoaniline+nano2+hcl
Given the model from the question,
- The products are: N₂, H₂O and H₂
- The reactants are: H₂ and NO
- The limiting reactant is H₂
- The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂
<h3>Balanced equation </h3>
From the model given, we obtained the ffolowing
- Red => Oxygen
- Blue => Nitrogen
- White => Hydrogen
Thus, we can write the balanced equation as follow:
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
- Reactants: H₂ and NO
- Product: N₂, H₂O and H₂
<h3>How to determine the limiting reactant</h3>
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
3 moles of H₂ reacted with 2 moles of NO.
Therefore,
5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO
From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.
Thus, H₂ is the limiting reactant
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Answer:
Ionic character
A. PF₃ > PBr₃ > PCl₃
B. BF₃ > CF₄ > NF₃
C. TeF₄ > BrF₃ > SeF₄
Explanation:
The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.
- Atoms with high electronegativity tend to form negative ions.
- Ionic compounds formed between elements with high electronegativity difference.
- % ionic character is directly proportional to electronegativity difference.
- According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
- ΔE.N (Electronegativity difference) between( P and F = 4 - 2.19 = 1.81), (P and Br = 2.96 - 2.19 = 0.77) , (P and Cl = 3.16 - 2.96 = 0.2 )
- ΔE.N (Electronegativity difference) between( N and F = 4 - 3 = 1), (B and F = 4 - 2 = 2) , (C and F = 4 - 2.5 = 1.5 )
- ΔE.N (Electronegativity difference) between( Se and F = 4 - 2.55 = 1.45), (F and Te = 4 - 2.1 = 1.9) , (F and Br = 4 - 2.19 = 1.81 )
The glass opposite to the negative electrode started to glow. Hence, option B is correct.
<h3>What is a cathode ray tube?</h3>
A cathode-ray tube (CRT) is a specialized vacuum tube in which images are produced when an electron beam strikes a phosphorescent surface.
J.J. Thomson, through his famous Cathode ray experiment, proved that all atoms contain small negatively charged particles known as electrons. In the experiment, he applied electric voltage across a cathode ray tube. a fluorescent material coating was done on the positive side. When the voltage was applied, the positive side has glowing dots.
Hence, option B is correct.
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