Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
Answer: GeH4 (Germanium(IV) Hydride)
Explanation:
A Binary molecular compound Hydrogen and a Group 4A element which is more more acidic than SiH4 in aqueous solution is GeH4.
The pKa of GeH4;
= 25
Whilst that of SiH4
= 35
The lesser the pKa the higher the Ka which means more acidic.
Answer:
The molecule has a bent geometry
Explanation:
Let us look again at the principles of VSEPR theory. The shape of a molecule depends on the number of electron pairs that surround the valence shell of the central atom in the molecule.
Lone pairs distort the molecular geometry away from what is expected on the basis of VSEPR theory.
The molecule described in the question has the form AEX2. Two substituents and one lone pair form three electron domains around the central atom. The expected geometry is trigonal planar but the observed molecular geometry is bent because of the lone pairs present.
Answer:
a metal spoon left in boiling water
Explanation:
when the thermal energy is the energy contained within a system that is responsible for its temperature.
and when the thermal energy is can be determined by this formula:
q = M * C *ΔT
when q is the thermal energy
and M is the mass of water = 100 g
and C is the specific heat capacity of water = 4.18 joules/gram.°C
and T is the difference in Temperature = 50 °C
So by substitution:
∴ q = 100 g * 4.18 J/g.°C * 50
= 20900 J = 20.9 KJ
