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
Unmmmm i’m not quite understanding brhajsnsbfbfhrndnsnsjsmsbgbtieoou that but yes
1 mol = 6.02 * 10^23 atoms of carbon
x mol = 1.45 * 10^24 atoms of carbon
1/x =6.02*10^23 / 1.45 * 10^24
6.02 * 10^23 x = 1.45 * 10^24
x = 1.45 * 10^24 / 6.02 * 10^23
x = 2.41 mols of carbon
Answer:The best phrase which supports the eruption would be high-viscosity magma lava broken into fragments.
