Answer:
i) B
ii) D
Explanation:
<em>Bond length is determined by the size of the atoms involved and the bond order </em>
A) C-I
B) H-I
answer : H-I has the shortest bond length because H has an electronegativity value of 2.2 while C has an electronegativity value of 2.5 hence the bond between H-I is greater than C - I due the electronegativity difference between H-I is greater as well.
C) H-Cl
D) H-I
answer : H-Cl has the shortest bond length due the electronegativity difference between H-CI is greater as well.
Answer:
[1] sulphur trioxide
Explanation:
Isoelectronic species have the same number of valence electrons.
Valence electrons in nitrate (NO₃⁻):
5e- (N) + (3 x 6e-)(3xO) + 1e- (charge) = 24e-
Valence electrons in sulphur trioxide (SO₃):
6e- (S) + (3 x 6e-)(3xO) = 24e-
Valence electrons in sulphite (SO₃²⁻):
6e- (S) + (3 x 6e-)(3xO) + 2e- (charge) = 26e-
Valence electrons in phosphine (PH₃):
5e- (P) + (3 x 1e-)(3xH) = 8e-
Valence electrons in water (H₂O):
6e- (O) + (2 x 1e-)(2xH) = 8e-
Valence electrons in chlorite (ClO₂⁻):
7e- (Cl⁻) + (2 x 6e-)(2xO) + 1e- (charge) = 20e-
The only species isoelectronic with nitrate is sulphur trioxide. Both have trigonal planar geoemetry.
Answer:
Tree sap flows over the leaf and preserves it.
Explanation:
Amber would preserve the image.
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
