I believe you do 151.7= 4.18x (4.70) because you use the formula q=mcat
Hot/cold packs are used by athletes to minimize swelling of injuries such as muscle and joint sprains. ... The hot/cold pack is activated by breaking the seal on the pouch of water and shaking the pack vigorously. This action mixes the water with the chemical starting the exothermic or endothermic reaction.
Answer:
Explanation:
2NaNO₃ = 2NaNO₂ + O₂
NaNO₃ = NaNO₂ + 1/2 O₂
- 468 kJ -369 kJ 0 kJ
enthalpy of decomposition reaction
= - 369 - ( - 468 ) kJ
= 99 kJ / mol .
Answer:
Q = 802.6 J
Explanation:
Given data:
Specific heat capacity of water = 4.18 J/g.°C
Mass of water = 12 g
Initial temperature = 23°C
Final temperature = 39°C
Heat required = ?
Solution:
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 = 39°C - 23°C
ΔT = 16 °C
Q = 12 g× 4.18 J/g.°C × 16 °C
Q = 802.6 J
