If the liquid is at or above its flash point, the flame spread rate is fast, and the entire pool is engulfed within seconds. ... As the liquid temperature decreases, flame radiation must both heat the liquid to the flash point temperature and supply the heat of vaporization.
Plasmolysed because of osmosis as the salt solution has lower water potential than the cells of the stalk
Answer:
A flame test of a colorless solution gives a bright yellow color. When reacted with AgNO, a white precipitate forms that dissolves when HNO, is added.
Answer:
T₂ = 221.8 K
Explanation:
Given data:
Initial temperature = 57.9°C ( 57.9+273 =330.9 k)
Initial volume = 75.8 mL
Final volume = 50.8 mL
Final temperature = ?
Solution:
According to Charles's law,
V₁ /T₁= V₂/T₂
T₂ = V₂T₁/V₁
T₂ = 50.8 mL ×330.9 k / 75.8 mL
T₂ = 16809.72 mL.K/ 75.8 mL
T₂ = 221.8 K
Answer : The amount of heat absorbed are, 258485.5 J
Solution :
Formula used :
where,
Q = heat gained = ?
m = mass of water = 650 g
c = specific heat of water = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get the final temperature of copper.
Therefore, the amount of heat absorbed are, 258485.5 J
