Answer:
5 kJ/g
Explanation:
There are two energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the solution
q₁ + q₂ = 0
m₁ΔH + m₂CΔT = 0
Data:
m₁ = 0.258 g
V₂ = 100 mL
C = 4.184 J°C⁻¹g⁻¹
T_i = 22 °C
T_f = 25.1 °C
Calculations
(a) Mass of solution
(b) ΔT
ΔT = T_f - T_i = 25.1 °C - 22 °C = 3.1°C
(c) ΔH
Note: The answer can have only one significant figure because you measured the initial temperature of the water only to the nearest degree.
Answer:
25.39
Explanation:
Given parameters:
Abundance of X-25 = 80.5%
Abundance of X - 27 = 19.5%
Unknown:
Average atomic mass of X = ?
Solution:
The average atomic mass of X can be derived using the expression below:
Average atomic mass = (abundance x mass of X - 25) + (abundance x mass of X - 27)
Average atomic mass = (80.5% x 25) + (19.5% x 27) = 25.39
The crudos is h2o combined with c3o
Vapourization and condensation. the actual 1 shud be vapourization.
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s