The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
Explanation:
Some Rules Regarding Oxidation Numbers:
- Hydrogen has oxidation number of + 1 except in hydrides where it is -1
- Oxygen has oxidation number of -2 except in peroxides where it is -1
- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1
- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.
A. HI
Hydrogen has oxidation of + 1
Oxidation number of I:
1 + x = 0
x = -1
B. PBr3
Br has oxidation number of - 1
Oxidation number of Pb:
x + 3 (-1) = 0
x = + 3
C. KH
Hydrogen has oxidation of + 1
Oxidation number of K:
1 + x = 0
x = -1
D. H3PO4
Hydrogen has oxidation number of + 1
Oxygen has oxidation number of -2
Oxidation number of P:
3(1) + x + 4(-2) = 0
3 + x - 8 =0
x = 5
