Answer:
1 and 2
Trigonal planar
Tetrahedral
Trigonal Planar
Linear
Bent ( v- shape)
Explanation:
The highlighted atoms has a delocalized lone pair - 1 and 2
All sp2-hybridized carbon atoms have geometry- Trigonal planar
All sp3-hybridized carbon atoms have geometry - Tetrahedral
The nitrogen atom has geometry - Trigonal Planar
The oxygen atom of the C=O bond has geometry - Linear
and the other oxygen atom has geometry - Bent ( v- shape)
P.S - The correct question is -
Answer:
D.) H-O
Explanation:
Polarity is determined based on the difference in electronegativity of the atoms. The greater the difference, the more polar the bond. The general trend is that the atoms in the top-right corner of the periodic table are the most electronegative.
A.) is incorrect because H-H has no electronegativity difference, making it nonpolar.
B.) and C.) are incorrect because their electronegativity differences are not the greatest.
D.) is correct because the electronegativity difference between the H and O is the greatest.
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
Answer:
[H2] = 0.0692 M
[I2] = 0.182 M
[HI] = 0.826 M
Explanation:
Step 1: Data given
Kc = 54.3 at 430 °C
Number of moles hydrogen = 0.714 moles
Number of moles iodine = 0.984 moles
Number of moles HI = 0.886 moles
Volume = 2.40 L
Step 2: The balanced equation
H2 + I2 → 2HI
Step 3: Calculate Q
If we know Q, we know in what direction the reaction will go
Q = [HI]² / [I2][H2]
Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Q =(n(HI)²) /(nH2 *nI2)
Q = 0.886²/(0.714*0.984)
Q =1.117
Q<Kc This means the reaction goes to the right (side of products)
Step 2: Calculate moles at equilibrium
For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI
Moles H2 = 0.714 - X
Moles I2 = 0.984 -X
Moles HI = 0.886 + 2X
Step 3: Define Kc
Kc = [HI]² / [I2][H2]
Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Kc =(n(HI)²) /(nH2 *nI2)
KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))
X = 0.548
Step 4: Calculate concentrations at the equilibrium
[H2] = (0.714-0.548) / 2.40 = 0.0692 M
[I2] = (0.984 - 0.548) / 2.40 = 0.182 M
[HI] = (0.886+2*0.548) /2.40 = 0.826 M
