If I remember correctly, you would have to heat the reaction beaker over a burner..
I apologize if I'm wrong
Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.
PV = nRT
V, T and R are constants so they can be lumped together to a constant k.
P/n = k
P1/n1 = P2/n2
since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
This is a tough one but the answer is water
think of it this way the ocean is water
Answer:
It won't let me type this for some reason but here it is.
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
