Answer:
Foods that are basic can be identified by its bitter taste. Option C
The given reaction is A---> products
Trial-1: The initial concentration of A is 0.10 M when the initial rate is 0.015 M/s
Trial-2: The initial concentration of A is 0.40 M when the initial rate is 0.060 M/s
Let the order be m
Rate law for first trial will be:
Rate law for second trial will be:
Rate- 2/ Rate-1 :
So, m = 1
Therefore the order of the reaction is 1
1. When that water melts in the Spring, the salinity decreases.
2. When ocean water evaporates, the salinity increases.
3. When ocean water freezes, the salinity of water increases.
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:
Q = 96.6 j
Explanation:
Given data:
Heat required = ?
Initial temperature = 19°C
Final temperature = 33°C
Mass of disc = 3.0 g
Specific heat capacity = 2.3 J/g.°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 33°C - 19°C
ΔT = 14°C
Q = 3.0 g×2.3 J/g.°C × 14°C
Q = 96.6 j