Answer:
Pp, Pp, pp, pp, 50% chance at purple and 50% chance at white
Explanation:
Just took the test and it was 32grams. On my test it was Block B: 32 Grams
I'm not sure what the question is but 3 is an example of a prime number. I hope this help
Answer:
Specific heat of substance = 0.897 J/g.°C
Explanation:
Given data:
Specific heat of substance = ?
Mass of substance = 25.0 g
Heat absorbed = 493.4 J
Initial temperature = 12.0 °C
Final temperature = 34°C
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 = 34°C -12.0°C
ΔT = 22°C
493.4 J = 25 g ×c× 22°C
493.4 J = 550 g.°C ×c
c = 493.4 J / 550 g.°C
c = 0.897 J/g.°C
Answer:
At a given temperature, a system of particles can be considered as point masses (m) each moving at a certain translational velocity (v). The motion of these particles can be defined in terms of their average translational kinetic energy which is responsible for the heat transfer during molecular collisions and therefore the temperature of the system.
The kinetic temperature T is given in terms of the average translational kinetic energy as:
T = 2/3k(Kinetic energy)
T = 2/3k(1/2*m*v²)
where K = Boltzmann constant
Ans: C) Average translational kinetic energy
