Answer:
Zn2+ is colourless
Explanation:
We know that transition metal salts are usually coloured due to the possibility of d-d transition.
This d-d transition can only occur when there are vacant d-orbitals. The electronic configuration [Ar] 4s23d8 suggests the presence of vacant d-orbitals and the possibility of the compounds of Zn2+ being coloured.
However, the absence of colours in Zn2+ compounds shows that there is no d-d transition(electronic) spectra observed for Zn2+ because the d orbitals are completely filled. This means that the correct electronic configuration of the ion is [Ar] 3d10.
The answer is 69.048%.
Mass of potassium dichromate (K₂Cr₂O₇) is 294 g, which is a 100%.
Given mass of chromium is 52 g. So, the mass percent of chromium in the compound is 17.687%:
52 g : x% = 294 g : 100%
x = 52 ÷ 294 × 100 %
x = 17.687%
Given mass of potassium is 39 g. <span>So, the mass percent of potassium in the compound is 13.265%:
39 g : x% = 294 g : 100%
x = </span>39 ÷ <span>294 × 100 %
x = 13.265%
Therefore, </span>the mass percent of oxygen in the compound is <span>69.048%:
100% - </span>17.687% - <span>13.265% = 69.048%.</span>
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.
