Acceleration is the rate at which an object changes its speed. When a raindrop is falling down from the clouds due to gravity, the veloicty is pretty much fast, however air resistance comes in action and slows the droplet down... it gives it a constant speed until it hits the ground
Answers:
(a) 30.55 °C
(b) 298 K and 77°F
(c) 204.44 °C and 477.44 K
(d) -320.8 °F and -196 °C
Explanation:
Converting °C into °F;
°F = °C × 1.8 + 32
Converting °F into °C;
°C = °F - 32 ÷ 1,8
Converting °C into K;
K = °C + 273
Converting K into °C;
°C = K - 273
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
Answer: I believe that it's electrons please let me know if it was right or what it was!
Hope this helps!!! Good luck!!! ;)
Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.

Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:

Now we have to calculate the energy required.

where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ